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UVALive 6529 Eleven 区间dp

2017-04-21 14:18 323 查看
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
;
int x, sum, len, app[10];
int C

, d[10][L][M], g

, 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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