UVALive 6529 Eleven 区间dp
题目链接:点击打开链接
题意:
给定一个数,又一次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; }
posted on 2017-04-21 14:18 gavanwanggw 阅读(141) 评论(0) 编辑 收藏 举报