[编程题]电话号码分身
- 热度指数：23448 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
继MIUI8推出手机分身功能之后，MIUI9计划推出一个电话号码分身的功能：首先将电话号码中的每个数字加上8取个位，然后使用对应的大写字母代替
（"ZERO", "ONE", "TWO", "THREE",
"FOUR", "FIVE", "SIX", "SEVEN",
"EIGHT", "NINE"）， 然后随机打乱这些字母，所生成的字符串即为电话号码对应的分身。
输入描述:
第一行是一个整数T（1 ≤ T ≤ 100)表示测试样例数；接下来T行，每行给定一个分身后的电话号码的分身（长度在3到10000之间）。
输出描述:
输出T行，分别对应输入中每行字符串对应的分身前的最小电话号码（允许前导0）。
示例1
输入
4 EIGHT ZEROTWOONE OHWETENRTEO OHEWTIEGTHENRTEO
输出
0 234 345 0345
要先依次解码具有独一无二字符的数字，具有这样特点的数字有五个分别是FOUR(U),SIX(X),
TWO(W),EIGHT(G),ZERO(Z),可以根据独特字符的个数直接判断有多少个相应的数字，例如有
3个U那么就一定有3个FOUR...,解码完成这五个数字之后另外的数字也会由于这些数字的移除
而具有了独一无二的字符，这样的数字有FIVE(F)，THREE(T),FIVE找到之后，只有SEVEN含
有V,所以又可以依据V字符的个数解码SEVEN的个数，最后剩下的ONE和NINE也具有了自己的
标志性字符分别是ONE(O),NINE(I)，需要注意的是原始数字和最终出现的数字还有一个转换
的过程（加8取个位数），所以还要相应转换回去。最后要注意的是，要求每行字符串对应的
分身前的最小电话号码，不要傻傻的按照字典序排序，这样时间复杂度过高，对于这个问题
可以用桶排序，10个bucket分别用于统计0-9出现的次数，最终桶中存储的结果依次输出就是
所有组合中最小的数字。
#include <iostream>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
int i = 0, j = 0;
int n;
while (cin >> n){
string s;
for (i = 0; i < n; i++){
cin >> s;
vector<int> iimap(256);
for (j = 0; j < s.size(); j++){
iimap[s[j]]++;
}
vector<int> res(10);
int count = iimap['U'];
res[6] = count;
iimap['F'] -= count;
iimap['O'] -= count;
iimap['U'] -= count;
iimap['R'] -= count;
count = iimap['X'];
res[8] = count;
iimap['S'] -= count;
iimap['I'] -= count;
iimap['X'] -= count;
count = iimap['W'];
res[4] = count;
iimap['T'] -= count;
iimap['W'] -= count;
iimap['O'] -= count;
count = iimap['G'];
res[0] = count;
iimap['E'] -= count;
iimap['I'] -= count;
iimap['G'] -= count;
iimap['H'] -= count;
iimap['T'] -= count;
count = iimap['Z'];
res[2] = count;
iimap['Z'] -= count;
iimap['E'] -= count;
iimap['R'] -= count;
iimap['O'] -= count;
count = iimap['F'];
res[7] = count;
iimap['F'] -= count;
iimap['I'] -= count;
iimap['V'] -= count;
iimap['E'] -= count;
count = iimap['V'];
res[9] = count;
iimap['S'] -= count;
iimap['E'] -= count;
iimap['V'] -= count;
iimap['E'] -= count;
iimap['N'] -= count;
count = iimap['T'];
res[5] = count;
iimap['T'] -= count;
iimap['H'] -= count;
iimap['R'] -= count;
iimap['E'] -= count;
iimap['E'] -= count;
count = iimap['O'];
res[3] = count;
iimap['O'] -= count;
iimap['N'] -= count;
iimap['E'] -= count;
count = iimap['I'];
res[1] = count;
iimap['N'] -= count;
iimap['I'] -= count;
iimap['N'] -= count;
iimap['E'] -= count;
for (int k = 0; k < res.size(); k++){
for (int l = 0; l < res[k]; l++)
cout << k;
}
cout << endl;
}
}
return 0;
}
编辑于 2017-08-24 15:49:06
回复(18)
**，**题
import java.util.Scanner;
/**
* Created by Evergreen on 2017/8/10.
*/
public class Main {
public static void handle(String str){
str=str.toLowerCase();
int[] hash=new int[10];
StringBuffer sb=new StringBuffer(str);
while(sb.toString().contains("z")){//zero
//2
hash[2]++;
sb.deleteCharAt(sb.indexOf("z"));
sb.deleteCharAt(sb.indexOf("e"));
sb.deleteCharAt(sb.indexOf("r"));
sb.deleteCharAt(sb.indexOf("o"));
}
while(sb.toString().contains("x")){//six
hash[8]++;
sb.deleteCharAt(sb.indexOf("s"));
sb.deleteCharAt(sb.indexOf("i"));
sb.deleteCharAt(sb.indexOf("x"));
}
while(sb.toString().contains("s")){//seven
hash[9]++;
sb.deleteCharAt(sb.indexOf("s"));
sb.deleteCharAt(sb.indexOf("e"));
sb.deleteCharAt(sb.indexOf("v"));
sb.deleteCharAt(sb.indexOf("e"));
sb.deleteCharAt(sb.indexOf("n"));
}
while(sb.toString().contains("u")){//four
hash[6]++;
sb.deleteCharAt(sb.indexOf("f"));
sb.deleteCharAt(sb.indexOf("o"));
sb.deleteCharAt(sb.indexOf("u"));
sb.deleteCharAt(sb.indexOf("r"));
}
while(sb.toString().contains("f")){//five
hash[7]++;
sb.deleteCharAt(sb.indexOf("f"));
sb.deleteCharAt(sb.indexOf("i"));
sb.deleteCharAt(sb.indexOf("v"));
sb.deleteCharAt(sb.indexOf("e"));
}
while(sb.toString().contains("g")){//eight
hash[0]++;
sb.deleteCharAt(sb.indexOf("e"));
sb.deleteCharAt(sb.indexOf("i"));
sb.deleteCharAt(sb.indexOf("g"));
sb.deleteCharAt(sb.indexOf("h"));
sb.deleteCharAt(sb.indexOf("t"));
}
while(sb.toString().contains("w")){//two
hash[4]++;
sb.deleteCharAt(sb.indexOf("t"));
sb.deleteCharAt(sb.indexOf("w"));
sb.deleteCharAt(sb.indexOf("o"));
}
while(sb.toString().contains("h")&&!sb.toString().contains("g")){//three
hash[5]++;
sb.deleteCharAt(sb.indexOf("t"));
sb.deleteCharAt(sb.indexOf("h"));
sb.deleteCharAt(sb.indexOf("r"));
sb.deleteCharAt(sb.indexOf("e"));
sb.deleteCharAt(sb.indexOf("e"));
}
while(sb.toString().contains("o")&&!sb.toString().contains("z")){//one
hash[3]++;
sb.deleteCharAt(sb.indexOf("o"));
sb.deleteCharAt(sb.indexOf("n"));
sb.deleteCharAt(sb.indexOf("e"));
}
while(sb.toString().contains("n")) {//nine
hash[1]++;
sb.deleteCharAt(sb.indexOf("n"));
sb.deleteCharAt(sb.indexOf("i"));
sb.deleteCharAt(sb.indexOf("n"));
sb.deleteCharAt(sb.indexOf("e"));
}
for(int i=0;i<10;i++){
for(int j=1;j<=hash[i];j++){
System.out.print(i);
}
}
System.out.println();
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNextInt()){
int n=sc.nextInt();
for(int i=0;i<n;i++)
handle(sc.next());
}
}
}
发表于 2017-08-10 14:23:11
回复(18)
思路基本是一楼的思路 但是他的代码太长了
实际上0 2 4 6 8这几个数字可以通过Z W U X G唯一确定
剩下的1 3 5 7可以通过O T F S唯一确定
而剩下的9随便哪个都可以确定 姑且用I
将这些按顺序写成一个表
这些都可以打表来简化代码 代码如下
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
string a[10] = {"ZERO", "TWO", "FOUR", "SIX", "EIGHT",
"ONE", "THREE", "FIVE", "SEVEN",
"NINE"};
int num_map[] = {0, 2, 4, 6, 8, 1, 3, 5, 7, 9};
char dic[] = {'Z', 'W', 'U', 'X', 'G', 'O', 'T', 'F', 'S', 'I'};
int ha[26];
vector<int> ans;
int main(){
int T; scanf("%d", &T);
while(T--){
string s; cin>>s;
ans.clear();
memset(ha, 0, sizeof(ha));
for(int i = 0; i < s.length(); i++) ha[s[i]-'A']++;
for(int i = 0; i < 10; i++){
int times = ha[dic[i]-'A'];
if(times > 0){
for(int j = 0; j < a[i].size(); j++)
ha[a[i][j]-'A'] -= times;
for(int k = times; k > 0; k--)
ans.push_back((num_map[i]+2) % 10);
}
}
sort(ans.begin(), ans.end());
for(int i=0;i<ans.size();i++) printf("%d", ans[i]);
printf("\n");
}
}
编辑于 2018-10-03 20:00:55
回复(0)
统计拥有的字母数量，然后可以找到规律，只有0（ZERO）有字母Z，也就是说，有多少个Z就有多少个0。同理，只有2有字母W，4有字母U，6有字母X，8有字母G。
将上述单词从字母统计中删去，发现剩下的单词中，只有5有字母F，1有字母O，3有字母R，7有字母S。
再删去，发现只有9有字母I。
将以上数字统计后排序可得。
import java.util.*;
import java.lang.*;
public class Main {
static String[] table = { "ZERO", "ONE", "TWO", "THREE", "FOUR", "FIVE",
"SIX", "SEVEN", "EIGHT", "NINE" };
static int retL = 0;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int i, j;
int n = sc.nextInt();
sc.nextLine();
String str;
int[] mark = new int[26];
int[] ret = new int[10000];
int temp = 0;
while (sc.hasNext()) {
retL = 0;
str = sc.nextLine();
for(i=0;i<str.length();i++) {
mark[str.charAt(i)-'A']++;
}
if(mark['Z' - 'A'] != 0) handle(mark, ret, 2, 'Z');
if(mark['W' - 'A'] != 0) handle(mark, ret, 4, 'W');
if(mark['U' - 'A'] != 0) handle(mark, ret, 6, 'U');
if(mark['X' - 'A'] != 0) handle(mark, ret, 8, 'X');
if(mark['G' - 'A'] != 0) handle(mark, ret, 0, 'G');
if(mark['F' - 'A'] != 0) handle(mark, ret, 7, 'F');
if(mark['O' - 'A'] != 0) handle(mark, ret, 3, 'O');
if(mark['R' - 'A'] != 0) handle(mark, ret, 5, 'R');
if(mark['S' - 'A'] != 0) handle(mark, ret, 9, 'S');
if(mark['I' - 'A'] != 0) handle(mark, ret, 1, 'I');
Arrays.sort(ret,0,retL);
for (i = 0; i < retL; i++) {
System.out.print(ret[i]);
}
System.out.println();
}
}
public static void handle(int[] mark, int[] ret, int num, char point) {
int temp = mark[point - 'A'],i;
for (i = 0; i < temp; i++) {
ret[retL++] = num;
}
for (i = 0; i < table[(num+8)%10].length(); i++) {
mark[table[(num+8)%10].charAt(i) - 'A'] -= temp;
}
}
}
编辑于 2017-08-09 05:11:54
回复(1)
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main { public static void main(String[] args) { // TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int num=Integer.parseInt(sc.nextLine());//行数
String[] sArr=new String[num];//构建一个String数组存放输入的每行字符串
for(int i=0;i<num;i++) {
sArr[i]=sc.nextLine();
}
for(String each:sArr) {
//构建一个map用于存储代表性字母的个数
//0-->Z; 2-->W; 4-->U; 6-->X; 8-->G;(唯一对应，每个字母的个数就代表其数字的个数)
//1-->O(1的个数计算方法：num(O)-num(0)-num(2)-num(4)= map(O)-map(Z)-map(W)-map(U))
//3-->R(3的个数计算方法：num(R)-num(0)-num(4)= map(R)-map(Z)-map(U))
//5-->F(5的个数计算方法：num(F)-num(4)= map(F)-map(U))
//7-->S(7的个数计算方法：num(S)-num(6)= map(S)-map(X))
//9-->N(注意，9里面有两个N,故计算结果应除以2：(num(N)-num(1)-num(7))/2=(map(N)-count(1)-count(7))/2)
Map<Character,Integer> map=new HashMap<Character,Integer>() {
{
put('Z',0);
put('W',0);
put('U',0);
put('X',0);
put('G',0);
put('O',0);
put('R',0);
put('F',0);
put('S',0);
put('N',0); }
};
for(int i=0;i<each.length();i++) {//统计当前字符串中有代表性字母的个数
char c=each.charAt(i);
if(map.containsKey(c))
map.put(c, 1+map.get(c));
else
continue;//遇到不在字典里的字母就跳过
}
int[] count=new int[10];
count[0]=map.get('Z');
count[2]=map.get('W');
count[4]=map.get('U');
count[6]=map.get('X');
count[8]=map.get('G');
count[1]=map.get('O')-map.get('Z')-map.get('W')-map.get('U');
count[3]=map.get('R')-map.get('Z')-map.get('U');
count[5]=map.get('F')-map.get('U');
count[7]=map.get('S')-map.get('X');
count[9]=(map.get('N')-count[1]-count[7])/2;//NINE中有两个N
int[] countnew=new int[10];
for(int i=0;i<10;i++){
countnew[i]=count[(i+8)%10];//使用countnew数组存储加8以前的个数
}
for(int i=0;i<10;i++) {//按照从0到9的顺序输出
for(int j=0;j<countnew[i];j++) {//每个数字输出的次数
System.out.print(i);
}
}
System.out.println();//输出完一个字符串换行
} }
}
发表于 2018-07-26 09:04:35
回复(0)
#include<stdio.h>
#include<string.h>
int v[10]={2,4,6,8,0,3,5,7,9,1},n,i,j,hash[10],tmp[26],cnt[26];
char dst[10][10]={"ZERO","TWO","FOUR","SIX",
"EIGHT","ONE","THREE","FIVE","SEVEN","NINE"},s[10005];
void dfs(int *x){
int i,cnt=0,j;
for(i=0;i<26;i++)
if(x[i]==0) cnt++;
if(cnt==26){
for(i=0;i<10;i++)
for(j=0;j<hash[i];j++) printf("%d",i); printf("\n");
return;
}
for(i=0;i<26;i++) tmp[i]=x[i];
for(i=0;i<10;i++){
int flag=1;
for(j=0;j<dst[i][j]!='\0';j++){
x[dst[i][j]-'A']--;
if(x[dst[i][j]-'A']<0){
flag=0;break;
}
}
if(flag==1){
hash[v[i]]++,dfs(x);break;
}
for(j=0;j<26;j++) x[j]=tmp[j];
}
}
int main(){
for(scanf("%d",&n);n--;dfs(cnt),memset(hash,0,sizeof(hash))){
scanf("%s",s);
memset(cnt,0,sizeof(cnt));
for(j=0;s[j]!='\0';j++) cnt[s[j]-'A']++;
}
}//dfs删除，注意删除顺序，0 2 4 6 8是具备唯一特征字母的
//排除0 2 4 6 8,之后，1 3 5 7是具备唯一特征的，最后就剩9了
编辑于 2017-10-23 08:49:06
回复(0)
var n=readline();
while(str=readline()){ changedNum(str);
}
function changedNum(str){
var reg1=[/Z/g,/W/g,/U/g,/X/g,/G/g];//0,2,4,6,8;
var reg2=[/O/g,/H/g,/F/g,/S/g,/N/g];//1,3,5,7,9;
var result=[];
var num_arr=new Array(10).fill(0);
for(let i=0,j=0;i<reg1.length;i++){ var num=str.match(reg1[i]); if(num){ num_arr[i+j]=num.length; } j++;
}
for(let i=0,j=1;i<reg2.length;i++){ var num=str.match(reg2[i]); switch(i){ case 0:{//1="O"-0-2-4 if(num){ num_arr[i+j]=num.length-num_arr[0]-num_arr[2]-num_arr[4]; } break; } case 1:{//3="H"-8 if(num){ num_arr[i+j]=num.length-num_arr[8]; } break; } case 2:{//5="F"-4 if(num){ num_arr[i+j]=num.length-num_arr[4]; } break; } case 3:{//7="S"-6 if(num){ num_arr[i+j]=num.length-num_arr[6]; } break; } case 4:{//9=("N"-1-7)/2 if(num){ num_arr[i+j]=(num.length-num_arr[1]-num_arr[7])/2; } break; } default:
break; } j++;
}
for(let i=0;i<num_arr.length;i++){ for(let j=0;j<num_arr[i];j++){ if(i>=8){ result.push(i-8); }else{ result.push(i+2); } } }
console.log(result.sort((x,y)=>{return x-y}).join(""));
}
发表于 2018-07-31 19:29:54
回复(0)
一开始忘记“’NINE”是两个N了，9的时候在数N /(ㄒoㄒ)/~~
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
char ch[] = { 'Z', 'W', 'X', 'S', 'U', 'O', 'R', 'F', 'G', 'I' };
int dig[] = { 0, 2, 6, 7, 4, 1, 3, 5, 8, 9 };
string ch_all[] = { "ZERO", "TWO", "SIX", "SEVEN", "FOUR", "ONE", "THREE", "FIVE", "EIGHT", "NINE" };
int T;
for (auto &i : dig)
{
i-= 8;
if (i < 0)
i+= 10;
}
while (cin >> T)
{
while (T--)
{
int count[26] = { 0 };
string str;
cin >> str;
vector<int> dig_cout;
for (auto i : str)
{
count[i - 'A']++;
}
for (int i = 0; i < 10; i++)
{
int tmp_c = count[ch[i] - 'A'];
if (tmp_c>0)
for (auto j : ch_all[i])
count[j - 'A'] -= tmp_c;
while (tmp_c--)
dig_cout.push_back(dig[i]);
}
sort(dig_cout.begin(), dig_cout.end());
for (auto i : dig_cout)
cout << i;
cout << endl;
}
}
return 0;
}
发表于 2017-08-21 17:32:48
回复(1)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;cin>>n;
while(n--)
{
string s;
cin>>s;
vector<int> v;
int a[10]={0};
while(s.find('Z')!=s.npos)
{
s.erase(s.find('Z'),1);
s.erase(s.find('E'),1);
s.erase(s.find('R'),1);
s.erase(s.find('O'),1);
a[2]++;
}
while(s.find('G')!=s.npos)
{
s.erase(s.find('G'),1);
s.erase(s.find('E'),1);
s.erase(s.find('I'),1);
s.erase(s.find('H'),1);
s.erase(s.find('T'),1);a[0]++;
}
while(s.find('X')!=s.npos)
{
s.erase(s.find('X'),1);
s.erase(s.find('I'),1);
s.erase(s.find('S'),1); a[8]++;
}
while(s.find('S')!=s.npos)
{
s.erase(s.find('V'),1);
s.erase(s.find('E'),1);
s.erase(s.find('E'),1);
s.erase(s.find('S'),1);
s.erase(s.find('N'),1);a[9]++;
}
while(s.find('W')!=s.npos)
{
s.erase(s.find('W'),1);
s.erase(s.find('T'),1);
s.erase(s.find('O'),1);a[4]++;
}
while(s.find('U')!=s.npos)
{
s.erase(s.find('F'),1);
s.erase(s.find('U'),1);
s.erase(s.find('O'),1);
s.erase(s.find('R'),1);a[6]++;
}
while(s.find('F')!=s.npos)
{
s.erase(s.find('F'),1);
s.erase(s.find('I'),1);
s.erase(s.find('V'),1);
s.erase(s.find('E'),1);a[7]++;
}
while(s.find('T')!=s.npos)
{
s.erase(s.find('T'),1);
s.erase(s.find('H'),1);
s.erase(s.find('R'),1);
s.erase(s.find('E'),1);
s.erase(s.find('E'),1);a[5]++;
}
while(s.find('I')!=s.npos)
{
s.erase(s.find('N'),1);
s.erase(s.find('I'),1);
s.erase(s.find('N'),1);
s.erase(s.find('E'),1);a[1]++;
}
a[3]=s.length()/3;
for(int i=0;i<10;i++)
v.insert(v.begin(),a[i],i );
sort(v.begin(), v.end());
for(int i=0;i<v.size();i++) cout<<v[i];
cout<<endl;
}
}
很LOW 的题，只要复制粘贴的快，再长也是一分钟的事。
发表于 2022-04-06 20:54:00
回复(0)
#coding=utf-8
n=int(input())
di=[[['Z','ZERO',2],['W','TWO',4],['U','FOUR',6],['X','SIX',8],['G','EIGHT',0]], #查询字典，分2部分，'ZERO','TWO'等包含特定唯一字符'z','w'
[['O','ONE',3],['R','THREE',5],['F','FIVE',7],['S','SEVEN',9]]] #'ONE','THREE'去掉上面的唯一字符后，也只包含特定唯一字符'O','R'
for i in range(n): #最后只剩'NINE'需要特殊处理
num=list(input()) #输入
num.sort() #排序
re=[]
for j in range(2): #调用字典
for d in di[j]: #字典内部遍历
if d[0] in num: #是否包含特定唯一字符
k=num.count(d[0]) #唯一特定字符个数就是代表数字个数，也是其他字母个数，如有k个'Z'就有k个'ERO'
for d0 in d[1]:
s=num.index(d0)
del num[s:s+k] #删除k个'ZERO'
re+=[d[2]]*k #代表的数字(减去8）
re+=[1]*(int(len(num)/4)) #剩余都是NINE
re.sort()#排序，找到最小的
re=[str(i) for i in re]
re=''.join(re)#组合
print(re)
n=int(input())
di=[[['Z','ZERO',2],['W','TWO',4],['U','FOUR',6],['X','SIX',8],['G','EIGHT',0]], #查询字典，分2部分，'ZERO','TWO'等包含特定唯一字符'z','w'
[['O','ONE',3],['R','THREE',5],['F','FIVE',7],['S','SEVEN',9]]] #'ONE','THREE'去掉上面的唯一字符后，也只包含特定唯一字符'O','R'
for i in range(n): #最后只剩'NINE'需要特殊处理
num=list(input()) #输入
num.sort() #排序
re=[]
for j in range(2): #调用字典
for d in di[j]: #字典内部遍历
if d[0] in num: #是否包含特定唯一字符
k=num.count(d[0]) #唯一特定字符个数就是代表数字个数，也是其他字母个数，如有k个'Z'就有k个'ERO'
for d0 in d[1]:
s=num.index(d0)
del num[s:s+k] #删除k个'ZERO'
re+=[d[2]]*k #代表的数字(减去8）
re+=[1]*(int(len(num)/4)) #剩余都是NINE
re.sort()#排序，找到最小的
re=[str(i) for i in re]
re=''.join(re)#组合
print(re)
发表于 2020-08-05 22:47:27
回复(0)
def func(x): #计算乱序字母中的ZERO,ONE,TWO...NINE的数量
dict = [ 0 for i in range(10)]
tab = [ 0 for i in range(10)]
num_Z = x.count('Z') #0 ZERO
num_W = x.count('W') #2 TWO
num_U = x.count('U') #4 FOUR
num_X = x.count('X') #6 SIX
num_G = x.count('G') #8 EIGHT
num_S = x.count('S') - num_X # 7 SEVEN
num_F = x.count('F') - num_U # 5 FIVE
num_T = x.count('T') - num_W - num_G # 3 THREE
num_O = x.count('O') - num_Z - num_W - num_U # 1 ONE
num_I = x.count('I') - num_F - num_X - num_G # 9 NINE
dict[0] = num_Z
dict[1] = num_O
dict[2] = num_W
dict[3] = num_T
dict[4] = num_U
dict[5] = num_F
dict[6] = num_X
dict[7] = num_S
dict[8] = num_G
dict[9] = num_I
tab[0] = dict[8]
tab[1] = dict[9]
tab[2] = dict[0]
tab[3] = dict[1]
tab[4] = dict[2]
tab[5] = dict[3]
tab[6] = dict[4]
tab[7] = dict[5]
tab[8] = dict[6]
tab[9] = dict[7]
s = ""
for i in range(len(tab)):
s += str(i)*tab[i]
return s
import sys
x = sys.stdin.readlines()
for i in range(1,len(x)):
print(func(x[i]))
编辑于 2018-09-05 10:38:15
回复(0)
//特别要注意9 nine 不能用n去判断除非除以2
import java.io.BufferedReader;
import java.io.IOException;import java.io.InputStreamReader;
/**
* @author Y.bear
* @version 创建时间：2018年9月3日 下午2:54:32 类说明
*/
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int times = Integer.valueOf(reader.readLine());
String[] results=new String[times];
for (int i = 0; i <times; i++) {
int[] result = new int[10];
String readLine = reader.readLine();
transform(readLine, result);
print(result,i,results);
}
for (String string : results) {
System.out.println(string);
}
}
public static void print(int[] result,int index,String[] results) {
StringBuffer buffer=new StringBuffer();
for (int i = 0; i < result.length; i++) {
while (result[i] > 0) {
result[i]--;
buffer.append(i);
}
}
results[index]=buffer.toString();
}
public static void getResult(int num, int count, int[] result) {
while (count > 0) {
count--;
result[(num + 2) % 10]++;
}
}
public static void transform(String line, int[] result) {
char[] letters = new char[26];
char[] charArray = line.toUpperCase().toCharArray();
for (char c : charArray) {
letters[c - 'A']++;
}
int zero = letters['Z' - 'A'];
getResult(0, zero, result);
letters['Z' - 'A'] -= zero;
letters['E' - 'A'] -= zero;
letters['R' - 'A'] -= zero;
letters['O' - 'A'] -= zero;
int two = letters['W' - 'A'];
getResult(2, two,result);
letters['T' - 'A'] -= two;
letters['W' - 'A'] -= two;
letters['O' - 'A'] -= two;
int four = letters['U' - 'A'];
getResult(4, four, result);
letters['F' - 'A'] -= four;
letters['O' - 'A'] -= four;
letters['U' - 'A'] -= four;
letters['R' - 'A'] -= four;
int six = letters['X' - 'A'];
getResult(6, six, result);
letters['S' - 'A'] -= six;
letters['I' - 'A'] -= six;
letters['X' - 'A'] -= six;
int eight = letters['G' - 'A'];
getResult(8, eight, result);
letters['E' - 'A'] -= eight;
letters['I' - 'A'] -= eight;
letters['G' - 'A'] -= eight;
letters['H' - 'A'] -= eight;
letters['T' - 'A'] -= eight;
int one = letters['O' - 'A'];
getResult(1, one, result);
letters['O' - 'A'] -= one;
letters['N' - 'A'] -= one;
letters['E' - 'A'] -= one;
int three = letters['T' - 'A'];
getResult(3, three, result);
letters['T' - 'A'] -= three;
letters['H' - 'A'] -= three;
letters['R' - 'A'] -= three;
letters['E' - 'A'] -= three;
letters['E' - 'A'] -= three;
int five = letters['F' - 'A'];
getResult(5, five, result);
letters['F' - 'A'] -= five;
letters['I' - 'A'] -= five;
letters['V' - 'A'] -= five;
letters['E' - 'A'] -= five;
int seven = letters['S' - 'A'];
getResult(7, seven, result);
letters['S' - 'A'] -= seven;
letters['E' - 'A'] -= seven;
letters['V' - 'A'] -= seven;
letters['E' - 'A'] -= seven;
letters['N' - 'A'] -= seven;
int nine = letters['I' - 'A'];
getResult(9, nine, result);
letters['N' - 'A'] -= nine;
letters['I' - 'A'] -= nine;
letters['N' - 'A'] -= nine;
letters['E' - 'A'] -= nine;
}
}
发表于 2018-09-03 16:48:21
回复(0)
//number代表字符串，bucket是转化后的数组
#include <iostream>
#include <string>
#include <map>
using namespace std;
int main(){
int n;
cin>>n;
while(n--){
string s;
cin >> s;
map <char,int> number;
int* bucket = new int [10];
for(int i = 0; i < s.length(); i++)
number[s[i]]++;
bucket[2] = number['Z'];
bucket[4] = number['W'];
bucket[6] = number['U'];
bucket[8] = number['X'];
bucket[0] = number['G'];
bucket[3] = number['O'] - bucket[2] - bucket[4] - bucket[6];
bucket[5] = number['R'] - bucket[2] - bucket[6];
bucket[7] = number['F'] - bucket[6];
bucket[9] = number['S'] - bucket[8];
bucket[1] = number['I'] - bucket[7] - bucket[8] - bucket[0];
for(int i = 0;i<10;i++){
while(bucket[i]--)
cout<<i;
}
cout<<endl;
}
return 0;
}
发表于 2018-04-17 20:39:17
回复(0)
//丧心病狂的题目 import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int t = Integer.parseInt(in.nextLine()); for (int i = 0; i < t; i++) { doWork(in); } } public static void doWork(Scanner in) { String s = in.nextLine(); int[] cs = new int[27]; for (int i = 0; i < s.length(); i++) { cs[s.charAt(i) - 'A']++; } int[] num = new int[10]; num[0] = cs['Z' - 'A']; num[6] = cs['X' - 'A']; num[2] = cs['W' - 'A']; num[4] = cs['U' - 'A']; num[8] = cs['G' - 'A']; num[3] = cs['T' - 'A'] - num[2] - num[8]; num[7] = cs['S' - 'A'] - num[6]; num[5] = cs['V' - 'A'] - num[7]; num[1] = cs['O' - 'A'] - num[0] - num[2] - num[4]; num[9] = (cs['N' - 'A'] - num[7] - num[1]) / 2; StringBuilder builder = new StringBuilder(); for (int i = 8; i <=9; i++) { for (int j = 0; j < num[i]; j++) { builder.append((i+2)%10); } } for (int i = 0; i <=7; i++) { for (int j = 0; j < num[i]; j++) { builder.append(i+2); } } System.out.println(builder.toString()); } }
发表于 2018-03-07 21:10:19
回复(0)
我的也是10%，我感觉测试数据给错了……
测试数据：
84
……
但是结果并没有84个测试数据，只有5个。
我的程序思路和大家大同小异：判断关键字。
比如ZERO,关键字为Z，那么数组a[10]内a[0]++，a[n]表示n的个数。
假如其他字母有O作为关键字，比如ONE，那么遇到Z，就执行：a[0]++,a[1]--；
a[1]--，就是要抵消掉ZERO里面的O给ONE带来的影响。
C++程序：
#include <iostream>
#include <string>
using namespace std;
//输入字符串str，返回对字符串总数字统计的结果a[10]
void check(int a[], string str);
//输出统计后的数组，比如a[0]=3,就输出3个0
void out_num(int a[10]);
int main() {
int num;
string str;
int a[100][10] = { 0 };
//网上测试，会用文件代替cin
while (cin>>num) {
for (int k = 0; k < num; k++) {
a[k][10] = { 0 };
cin >> str;
check(a[k], str);
}
for (int i = 0; i < num; i++) {
out_num(a[i]);
}
}
}
void out_num(int a[10]) {
//输出数字，8→0,9→1,0→2……
for (int k = 0; k < a[8]; k++)
cout << 0;
for (int k = 0; k < a[9]; k++)
cout << 1;
for (int j = 0; j < 8; j++)
for (int k = 0; k < a[j]; k++)
cout << j + 2;
cout<<endl;
}
void check(int a[], string str) {
int i = 0;
//假如用scanf，就要用\0判断结束标志，预留一个字节系统自动放置\0；
//假如用gets，就用\n判断；因为它会把你的\n也会读过来。
while (str[i] != '\0') {
switch (str[i]) {
case 'Z':
a[0]++;
a[1]--; //O
break;
case 'O':
a[1]++;
break;
case 'W':
a[2]++;
a[1]--; //O
break;
case 'H':
a[3]++;
break;
case 'U':
a[4]++;
a[1]--; //O
break;
case 'V':
a[5]++;
a[9]--; //I
break;
case 'X':
a[6]++;
a[7]--; //S
a[9]--; //I
break;
case 'S':
a[7]++;
a[5]--; //V
break;
case 'G':
a[8]++;
a[3]--; //H
a[9]--; //I
break;
case 'I':
a[9]++;
break;
default:
break;
}
i++;
}
}
测试数据：
84
……
但是结果并没有84个测试数据，只有5个。
我的程序思路和大家大同小异：判断关键字。
比如ZERO,关键字为Z，那么数组a[10]内a[0]++，a[n]表示n的个数。
假如其他字母有O作为关键字，比如ONE，那么遇到Z，就执行：a[0]++,a[1]--；
a[1]--，就是要抵消掉ZERO里面的O给ONE带来的影响。
C++程序：
#include <iostream>
#include <string>
using namespace std;
//输入字符串str，返回对字符串总数字统计的结果a[10]
void check(int a[], string str);
//输出统计后的数组，比如a[0]=3,就输出3个0
void out_num(int a[10]);
int main() {
int num;
string str;
int a[100][10] = { 0 };
//网上测试，会用文件代替cin
while (cin>>num) {
for (int k = 0; k < num; k++) {
a[k][10] = { 0 };
cin >> str;
check(a[k], str);
}
for (int i = 0; i < num; i++) {
out_num(a[i]);
}
}
}
void out_num(int a[10]) {
//输出数字，8→0,9→1,0→2……
for (int k = 0; k < a[8]; k++)
cout << 0;
for (int k = 0; k < a[9]; k++)
cout << 1;
for (int j = 0; j < 8; j++)
for (int k = 0; k < a[j]; k++)
cout << j + 2;
cout<<endl;
}
void check(int a[], string str) {
int i = 0;
//假如用scanf，就要用\0判断结束标志，预留一个字节系统自动放置\0；
//假如用gets，就用\n判断；因为它会把你的\n也会读过来。
while (str[i] != '\0') {
switch (str[i]) {
case 'Z':
a[0]++;
a[1]--; //O
break;
case 'O':
a[1]++;
break;
case 'W':
a[2]++;
a[1]--; //O
break;
case 'H':
a[3]++;
break;
case 'U':
a[4]++;
a[1]--; //O
break;
case 'V':
a[5]++;
a[9]--; //I
break;
case 'X':
a[6]++;
a[7]--; //S
a[9]--; //I
break;
case 'S':
a[7]++;
a[5]--; //V
break;
case 'G':
a[8]++;
a[3]--; //H
a[9]--; //I
break;
case 'I':
a[9]++;
break;
default:
break;
}
i++;
}
}
发表于 2018-02-05 16:13:22
回复(2)
提示格式错误，但是没找出来，有人能帮忙看看吗？
import java.util.Scanner;
public class Number {
public static void main(String[] argv) {
Scanner sin = new Scanner(System.in);
int count = sin.nextInt();
while (count-- >= 0) {
String str = sin.nextLine();
int[] stringNumber = new int[26];
String reString = "";
String[] array = { "EIGHT", "NINE", "ZERO", "ONE", "TWO", "THREE",
"FOUR", "FIVE", "SIX", "SEVEN" };
char[] strs = str.toCharArray();
for (int i = 0; i < strs.length - 1; i++) {
stringNumber[strs[i] - 'A'] += 1;
}
for (int i = 0; i < array.length; i++) {
reString = reString
+ dealString(stringNumber, array[i], i + "");
}
System.out.printf(reString + "\n");
}
}
private static String dealString(int[] stringNumber, String str, String tag) {
String string = "";
char[] numStrs = str.toCharArray();
for (int i = 0; i < numStrs.length - 1; i++) {
if (stringNumber[numStrs[i] - 'A'] <= 0) {
return string;
}
}
for (int i = 0; i < numStrs.length - 1; i++) {
stringNumber[numStrs[i] - 'A'] -= 1;
}
return tag + dealString(stringNumber, str, tag);
}
}
发表于 2018-01-30 23:13:30
回复(0)
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{ char ch[] = {'Z','W','X','S','U','O','R','F','G','I'}; int d[] = {0,2,6,7,4,1,3,5,8,9}; string num[] = {"ZERO","TWO","SIX","SEVEN","FOUR","ONE","THREE","FIVE","EIGHT","NINE"}; int T; for(int i=0;i<10;i++) { d[i] -= 8; if(d[i]<0) d[i] += 10; } while(cin>>T) { while(T--) { int count[26]={0}; string s; cin>>s; vector<int> result; for(int i=0;i<s.length();i++) count[s[i]-'A']++; for(int i=0;i<10;i++) { int t = count[ch[i]-'A']; if(t>0) for(int j=0;j<num[i].length();j++) count[num[i][j]-'A'] -= t; while(t--) result.push_back(d[i]); } sort(result.begin(), result.end()); for(int i=0;i<result.size();i++) cout<<result[i]; cout<<endl; } } return 0;
}
发表于 2018-01-17 01:19:33
回复(0)
#include
#include
#include
#include
#include
using namespace std;
typedef struct
{
int num;
char index;
string numword;
int numcnt;
}node;
void handle(string &numstr)
{
vector nodevec(10);
nodevec[0].num = 2;
nodevec[0].index = 'Z';
nodevec[0].numword = "ZERO";
nodevec[1].num = 4;
nodevec[1].index = 'W';
nodevec[1].numword = "TWO";
nodevec[2].num = 6;
nodevec[2].index = 'U';
nodevec[2].numword = "FOUR";
nodevec[3].num = 8;
nodevec[3].index = 'X';
nodevec[3].numword = "SIX";
nodevec[4].num = 0;
nodevec[4].index = 'G';
nodevec[4].numword = "EIGHT";
nodevec[5].num = 3;
nodevec[5].index = 'O';
nodevec[5].numword = "ONE";
nodevec[6].num = 5;
nodevec[6].index = 'T';
nodevec[6].numword = "THREE";
nodevec[7].num = 7;
nodevec[7].index = 'F';
nodevec[7].numword = "FIVE";
nodevec[8].num = 9;
nodevec[8].index = 'S';
nodevec[8].numword = "SEVEN";
nodevec[9].num = 1;
nodevec[9].index = 'E';
nodevec[9].numword = "NINE";
vector res;
for(vector::iterator iter = nodevec.begin();iter != nodevec.end();++iter)
{
node n = *iter;
n.numcnt = count_if(numstr.begin(), numstr.end(), bind2nd(equal_to(), n.index));
for(int k = 0;k < n.numcnt;k++)
{
for(string::iterator iter2 = n.numword.begin();iter2 != n.numword.end();++iter2)
{
numstr.erase(find_if(numstr.begin(), numstr.end(), bind2nd(equal_to(), *iter2)));
}
res.push_back(n.num);
}
}
sort(res.begin(), res.end(), less());
copy(res.begin(), res.end(), ostream_iterator(cout));
cout << endl;
}
int main()
{
int cnt;
while(cin >> cnt)
{
vector strvec(cnt);
for(int i = 0;i < cnt;i++)
{
cin >> strvec[i];
}
for(int i = 0;i < cnt;i++)
{
string numstr = strvec[i];
handle(numstr);
}
}
return 0;
}
时间：2018年8月14日
#include <iostream>
#include <string>
#include <map>
#include <vector>
using namespace std;
vector<int> func(string str) {
map<int, int> resmap;
map<char, int> amap;
for(int i = 0;i < str.size();i++) {
amap[str[i]]++;
}
//2
resmap[2] += amap['Z'];
for(int i = 0;i < resmap[2];i++) {
amap['Z']--;
amap['E']--;
amap['R']--;
amap['O']--;
}
resmap[4] += amap['W'];
for(int i = 0;i < resmap[4];i++) {
amap['T']--;
amap['W']--;
amap['O']--;
}
resmap[6] += amap['U'];
for(int i = 0;i < resmap[6];i++) {
amap['F']--;
amap['O']--;
amap['U']--;
amap['R']--;
}
resmap[8] += amap['X'];
for(int i = 0;i < resmap[8];i++) {
amap['S']--;
amap['I']--;
amap['X']--;
}
resmap[0] += amap['G'];
for(int i = 0;i < resmap[0];i++) {
amap['E']--;
amap['I']--;
amap['G']--;
amap['H']--;
amap['T']--;
}
resmap[3] += amap['O'];
for(int i = 0;i < resmap[3];i++) {
amap['O']--;
amap['N']--;
amap['E']--;
}
resmap[5] += amap['R'];
for(int i = 0;i < resmap[5];i++) {
amap['T']--;
amap['H']--;
amap['R']--;
amap['E']--;
amap['E']--;
}
resmap[7] += amap['F'];
for(int i = 0;i < resmap[7];i++) {
amap['F']--;
amap['I']--;
amap['V']--;
amap['E']--;
}
resmap[9] += amap['V'];
for(int i = 0;i < resmap[9];i++) {
amap['S']--;
amap['E']--;
amap['V']--;
amap['E']--;
amap['N']--;
}
resmap[1] += amap['I'];
for(int i = 0;i < resmap[1];i++) {
amap['N']--;
amap['I']--;
amap['N']--;
amap['E']--;
}
vector<int> vec;
for(auto i : resmap) {
for(int j = 0;j < i.second;j++){
vec.push_back(i.first);
}
}
return vec;
}
int main() {
int n;
cin >> n;
for(int i = 0;i < n;i++) {
string str;
cin >> str;
vector<int> vec = func(str);
for(auto num : vec) {
cout << num;
}
cout << endl;
}
return 0;
}
编辑于 2018-08-14 14:12:33
回复(0)
//找到数字和单词的对应关系，单词和字母的对应关系
import java.util.HashMap;
import java.util.Scanner;
/**
*
* @author JSZha0
*/
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
String[] sb = new String[n];
in.nextLine();
while(in.hasNext())
System.out.println(minPhoneNumber(in.nextLine()));
}
}
public static String minPhoneNumber(String str) {
//hashmap存储字符串中字母个数
HashMap<Character, Integer> map = new HashMap<>();
for(char i=65; i<91; i++) {
map.put(i, 0);
}
//统计字母个数
for(char s : str.toCharArray()) {
map.put(s, map.get(s)+1);
}
//数组对应关系，index表示数字，存储的值表示数字的个数
int[] ch = new int[10];
//
ch[2] = map.get('Z'); // ZERO 0 2
ch[4] = map.get('W'); // TWO 2 4
ch[6] = map.get('U'); // FOUR 4 6
ch[8] = map.get('X'); // SIX 6 8
ch[0] = map.get('G'); // EIGHT 8 0
ch[5] = map.get('H')-ch[0]; // THREE-GIGHT 3 5
ch[7] = map.get('F')-ch[6]; //...
ch[9] = map.get('V')-ch[7];
ch[3] = map.get('O')-ch[2]-ch[4]-ch[6];
ch[1] = map.get('I')-ch[0]-ch[8]-ch[7];
String s = "";
for(int i=0; i<10;i++) {
while(ch[i]>0) {
s += i;
ch[i]--;
}
}
return s;
}
}
发表于 2017-10-04 20:42:19
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