网易笔试编程（附ak代码）
1.十进制的和
#include <stdio.h>
#include <stack>
using namespace std;
int main() {
int n, num;
scanf("%d", &n);
while (n--) {
scanf("%d", &num);
stack<int> s;
while (num > 9) {
s.push(9);
num -= 9;
}
s.push(num);
while (!s.empty()) {
printf("%d", s.top());
s.pop();
}
printf("\n");
}
return 0;
}
2.ABpq（long long前90%）
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
while (n--) {
int ans = 1;
long long int a, b, p, q;
scanf("%lld%lld%lld%lld", &a, &b, &p, &q);
while (b - a > p) {
p *= q;
ans++;
}
printf("%d\n",ans);
}
return 0;
} 3.积木堆（long long 前30%）
#include <stdio.h>
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n;
long long int sum, now;
scanf("%d%lld", &n,&sum);
for (int i = 0; i < n; i++) {
scanf("%lld", &now);
if (sum >= 0) {
if (now + sum >= i) {
sum += now - i;
}
else {
sum = -1;
}
}
}
if (sum < 0) { printf("NO\n");
}
else {
printf("YES\n");
}
}
return 0;
} 4.跳木棍
判断vis[j]之前列了个表
- 若i到j不需要超能力
| vis[i] \ vis[j] | -1 | 0 | 1 | 2 |
| -1 | -1 | -1 | 1 | 2 |
| 1 | 1 | 1 | 1 | 1 |
| 2 | 2 | 2 | 1 | 2 |
- 若需要用到超能力
| vis[i] \ vis[j] | -1 | 0 | 1 | 2 |
| -1 | -1 | -1 | 1 | 2 |
| 1 | 2 | 2 | 1 | 2 |
| 2 | -1 | -1 | 1 | 2 |
#include <stdio.h>
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, k;
int h[1005], vis[1005];//vis为0表示未判断，为1表示可以跳到，为2表示超能力跳到，为-1表示不能跳到
scanf("%d%d", &n,&k);
for (int i = 0; i < n; i++) {
scanf("%d", &h[i]);
vis[i] = 0;
}
vis[0] = 1;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j <= i + k && j < n; j++) {
if (vis[j] == 1) continue;
if (h[j] <= h[i]) {
if (vis[j] < 1 || vis[i] == 1) {
vis[j] = vis[i];
}
else {
vis[j] = 2;
}
}
else {
if (vis[j] > 1 || vis[i] == 1) {
vis[j] = 2;
}
else {
vis[i] = -1;
}
}
}
}
if (vis[n - 1] > 0) {
printf("YES\n");
}
else {
printf("NO\n");
}
}
return 0;
} 
