250322网易互娱后端暑期笔试
贴个Java代码
求股票购买和售出的最大收益
贪心，如果这一天存在正数的收益，那么就直接买入，第二天售出即可。因为这样，就可以保证下一次，可以买到更多的股票
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextInt()) { // 注意 while 处理多个 case
int N = in.nextInt(), M = in.nextInt(), K = in.nextInt();
double[][] nums = new double[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
nums[i][j] = in.nextDouble();
}
}
double res = K;
int[][] records = new int[N][2];
for (int i = 0; i < N; i++) { // 遍历每一天的股票行情
int[] cur = new int[2];
if (i == 0) cur[0] = -1; // 第0天没有卖出
else cur[0] = records[i-1][1]; // 这一天的卖出，就是上一天的买入
double max = -(int)1e9;
int maxIdx = -1;
if (i == N-1) cur[1] = -1; // 最后一天没有买入
else {
for (int j = 0; j < M; j++) { // 找到这一天最大收益的股票
if (nums[i+1][j]/nums[i][j] > max) {
max = nums[i+1][j]/nums[i][j];
maxIdx = j;
}
}
if (max > 1) { // 大于1才有收益
cur[1] = maxIdx;
res *= max;
} else {
cur[1] = -1;
}
}
records[i] = cur;
}
System.out.println(String.format("%.4f", res));
for (int i = 0; i < N; i++) {
System.out.println(records[i][0] + " " + records[i][1]);
}
}
}
}
计算2048矩阵在上下左右移动后的结果矩阵
艰难模拟，写了好久。。
刚开始没考虑到中间有0值的情况，卡在9%，比如说 8 0 8 0 左移的结果是 16 0 0 0
import java.util.*;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextInt()) { // 注意 while 处理多个 case
int T = in.nextInt();
while (T-- > 0) {
int c = in.nextInt();
int[] ops = new int[c];
for (int i = 0; i < c; i++) ops[i] = in.nextInt();
int m = in.nextInt(), n = in.nextInt();
int[][] grid = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
grid[i][j] = in.nextInt();
}
}
for (int op : ops) {
if (op == 0) { // up
for (int j = 0; j < n; j++) {
int idx = 0; // 要更新写入的坐标
for (int i = 0; i < m; i++) {
if (grid[i][j] == 0) continue; // 跳过0值
int cur = grid[i][j]; // 当前不为0的值
while (i+1 < m && grid[i+1][j] == 0) i++; // 找到下一个不为0的值
if (i+1 < m && cur == grid[i+1][j]) { // 相等则合并
grid[idx++][j] = cur*2;;
i++;
} else { // 不相等则保存当前值到坐标即可
grid[idx++][j] = cur;
}
}
while (idx < m) grid[idx++][j] = 0;
// System.out.println(Arrays.toString(col));
}
} else if (op == 1) { // left
for (int i = 0; i < m; i++) {
int[] row = grid[i];
int idx = 0;
for (int j = 0; j < n; j++) {
if (row[j] == 0) continue;
int cur = row[j];
while (j+1 < n && row[j+1] == 0) j++;
if (j+1 < n && cur == row[j+1]) {
row[idx++] = cur*2;
j++;
} else {
row[idx++] = cur;
}
}
while (idx < n) row[idx++] = 0;
}
} else if (op == 2) { // down
for (int j = 0; j < n; j++) {
int idx = m-1;
for (int i = m-1; i >= 0; i--) {
if (grid[i][j] == 0) continue;
int cur = grid[i][j];
while (i-1 >= 0 && grid[i-1][j] == 0) i--;
if (i-1 >= 0 && cur == grid[i-1][j]) {
grid[idx--][j] = cur*2;
i--;
} else {
grid[idx--][j] = cur;
}
}
while (idx >= 0) grid[idx--][j] = 0;
}
} else { //right
for (int i = 0; i < m; i++) {
int[] row = grid[i];
int idx = n-1;
for (int j = n-1; j >= 0; j--) {
if (row[j] == 0) continue;
int cur = row[j];
while (j-1 >= 0 && row[j-1] == 0) j--;
if (j-1 >= 0 && cur == row[j-1]) {
row[idx--] = cur*2;
j--;
} else {
row[idx--] = cur;
}
}
while (idx >= 0) row[idx--] = 0;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
System.out.print(grid[i][j] + " ");
}
System.out.println();
}
}
}
}
}
a*x + b*y + c*z + d*w = n，求xyzw的可能取值，输出字典序最小的那个结果
xyzw的范围是0到2500
提前处理c*z + d*w的所有可能结果，用HashMap保存，然后遍历a和b查看是否有满足的情况
import java.util.*;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextInt()) { // 注意 while 处理多个 case
int a = in.nextInt(), b = in.nextInt(), c = in.nextInt(), d = in.nextInt(), n = in.nextInt();
HashMap<Long, int[]> map = new HashMap<>();
for (int i = 0; i <= 2500; i++) { // `c*z + d*w`的所有可能结果
for (int j = 0; j <= 2500; j++) {
map.putIfAbsent((long)i*c + (long)j*d, new int[]{i, j}); // 这里不能覆盖，才能保证c和d字典序最小
}
}
String res = "";
for (int i = 0; i <= 2500; i++) {
if (res.length() > 0) break;
for (int j = 0; j <= 2500; j++) {
long cur = (long)i*a + (long)j*b;
if (n-cur < 0) break; // 由于都大于0，所以更大的j更不行了，这里可以提前跳出
int[] val = map.get(n-cur);
if (val != null) { // 满足题目条件，保存结果
res = i + " " + j + " " + val[0] + " " + val[1];
break;
}
}
}
if (res.length() > 0) System.out.println(res);
else System.out.println(-1);
}
}
}
