UVALive 6529 Eleven 区间dp

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题目链接：点击打开链接

题意：

给定一个数，重新排列这个数的各个位置使得

1、无前导0

2、能被11整除

问：

有多少种组合方法

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1000000000 + 7;const int N = 100+2;const int L = 50+2;const int M = L*9;char s[N];int x, sum, len, app[10];int C[N][N], d[10][L][M], g[N][N], tot[12];int c(int x, int y) {if (~C[x][y])return C[x][y];if (x == 1 || y==0)return C[x][y] = 1;C[x][y] = 0;for (int i = 0; i <= y; ++i)C[x][y] = (C[x][y] + c(x-1, y-i))%mod;return C[x][y];}void work() {int v;len = strlen(s);sum = 0;memset(app, 0, sizeof app);for (int i = 0; i < len; ++i) {++ app[s[i]-'0'];sum += s[i]-'0';}tot[0] = 0;for (int i = 1; i <= 9; ++i)tot[i] = tot[i-1] + app[i];x = (len+1)/2;memset(d, 0, sizeof d);d[0][0][0] = 1;for (int i = 0; i <= 8; ++i)for (int j = 0; j <= x && j <= tot[i]; ++j)for (int s = 0; s <= j*9; ++s)if (d[i][j][s] > 0)for (int k = 0; k <= app[i+1] && k+j <= x; ++k) {v = (ll)d[i][j][s] * c(j+1,k) % mod;v = (ll)v*g[len-x-tot[i]+j][app[i+1]-k]%mod;d[i+1][j+k][s+k*(i+1)] = (d[i+1][j+k][s+k*(i+1)] + v)%mod;}int ans = 0;for (int j = 1; j <= x; ++j)for (int s = 0; s <= j*9; ++s)if (d[9][j][s] > 0 && abs(sum-s-s) % 11 == 0) {int k = x - j;if (k > app[0])continue;ans += (ll)d[9][j][s] * c(j,k) % mod;ans %= mod;}printf("%d\n", ans);}int main() {memset(C, -1, sizeof C);memset(g, 0, sizeof g);for (int i = 0; i < N; ++i) {g[i][0] = g[i][i] = 1;for (int j = 1 ; j < i; ++j)g[i][j] = (g[i-1][j] + g[i-1][j-1])%mod;}while (~scanf("%s", s))work();return 0;}

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