hdu 5389 Zero Escape（dp）

题目链接：hdu 5389 Zero Escape

注意全部分给一边的情况。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int mod = 258280327;const int maxn = 1e5 + 5;int N, A, B, dp[maxn][15], D[maxn];int main () {int cas;scanf("%d", &cas);while (cas--) {scanf("%d%d%d", &N, &A, &B);A %= 9, B %= 9;int s = 0;for (int i = 1; i <= N; i++) {scanf("%d", &D[i]);s = (s + D[i]) % 9;}if ((s - A + 9) % 9 != B) {int ans = 0;if (s == A)ans++;if (s == B)ans++;printf("%d\n", ans);continue;}memset(dp, 0, sizeof(dp));dp[0][0] = 1;for (int i = 1; i <= N; i++) {for (int j = 0; j < 10; j++) {dp[i][j] = (dp[i][j] + dp[i-1][j]) % mod;dp[i][(j+D[i])%9] = (dp[i][(j+D[i])%9] + dp[i-1][j]) % mod;}}printf("%d\n", dp[N][A]);}return 0;}