HDU 5464 ( Clarke and problem ) （dp）

dp[i][j] := 前i个数和为j的情况(mod p)

dp[i][j] 分两种情况 1.不选取第i个数 -> dp[i][j] = dp[i-1][j]

                          2.   选取第i个数 -> dp[i][j] = dp[i-1][t]　((t+a[i])%p==j)

(为什么很简单的题，思路也有了，比赛的时候就是写不对呢？)

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#include <iostream> #include <cstdio> #include <cstring> using namespace std;   typedef long long ll;   ll a[1005]; ll dp[1005][1005]; const ll MOD = 1000000007;   int main() {     int t, n, p;     scanf("%d", &t);     while (t--) {         memset(dp, 0, sizeof dp);         scanf("%d%d", &n, &p);           for (int i = 1; i <= n; ++i) {             scanf("%I64d", &a[i]);         }         dp[0][0] = 1;         for (int i = 1; i <= n; ++i) {             for (int j = 0; j < p; ++j) {                 int temp = (j - (a[i] % p) + p) % p;                 dp[i][j] = (dp[i - 1][temp] + dp[i - 1][j]) % MOD;             }         }         printf("%I64d\n", dp[n][0]);     }     return 0; }   /** Input: 78 4 2 1 2 3 4   Output: 8 */