UVALive 6529 Eleven 区间dp

标签：blog   http   io   ar   for   sp   on   2014   log   

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;
}

UVALive 6529 Eleven 区间dp

标签：blog   http   io   ar   for   sp   on   2014   log   

原文地址：http://blog.csdn.net/qq574857122/article/details/40618335