UVALive

题目链接：

http://acm.hust.edu.cn/vjudge/problem/47664

Eleven

Time Limit: 5000MS #### 问题描述 > In this problem, we refer to the digits of a positive integer as the sequence of digits required to write > it in base 10 without leading zeros. For instance, the digits of N = 2090 are of course 2, 0, 9 and 0. > Let N be a positive integer. We call a positive integer M an eleven-multiple-anagram of N if and > only if (1) the digits of M are a permutation of the digits of N, and (2) M is a multiple of 11. You are > required to write a program that given N, calculates the number of its eleven-multiple-anagrams. > As an example, consider again N = 2090. The values that meet the first condition above are 2009, > 2090, 2900, 9002, 9020 and 9200. Among those, only 2090 and 9020 satisfy the second condition, so > the answer for N = 2090 is 2.

输入

The input file contains several test cases, each of them as described below.
A single line that contains an integer N (1 ≤ N ≤ 10100).

输出

For each test case, output a line with an integer representing the number of eleven-multiple-anagrams
of N . Because this number can be very large, you are required to output the remainder of dividing it
by 109 + 7.

样例

sample input
2090
16510
201400000000000000000000000000

sample output
2
12
0

题意

给你一串数，求由这些数排列组合成的能被11整除的数的个数。

题解

能被11整除的数有一个特点，（偶数位的和-奇数位的和）%11=0;
dp[i][j][k] (0<=i-1<=9) 表示统计到i-1的时候，偶数位已经有j个数时,(y-x)%11==k(y代表偶数位和，x代表奇数位和)的情况数。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;

const int maxn = 111;
const int maxm = 15;
const int mod = 1e9 + 7;
char str[maxn];
int cntv[maxm];
int n;
LL C[maxn][maxn];

LL dp[maxm][maxn][maxm];
LL solve(int n) {
	memset(dp, 0, sizeof(dp));
	dp[0][0][0] = 1;
	int sum = 0,need=n/2;
	for (int i = 1; i <= 10; i++) {
		for (int j = 0; j <= sum&&j <= need; j++) {
			for (int dj = 0; dj <= cntv[i - 1] && (j + dj) <= need; dj++) {
				int dx = (dj - (cntv[i - 1] - dj))*(i-1)%11;
				dx = (dx%11 + 11) % 11;
				for (int k = 0; k < 11; k++) {
					int nk = (k + dx) % 11;
					dp[i][j + dj][nk] = (dp[i][j + dj][nk]+dp[i - 1][j][k]*C[need-j][dj]%mod*C[n-need-(sum-j)][cntv[i-1]-dj]%mod)%mod;
				}
			}
		}
		sum += cntv[i - 1];
	}
	return dp[10][need][0];
}

void pre() {
	memset(C, 0, sizeof(C));
	C[0][0] = 1;
	for (int i = 1; i < maxn; i++) {
		C[i][0] = 1;
		for (int j = 1; j <= i; j++) {
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
		}
	}
}

void init() {
	memset(cntv, 0, sizeof(cntv));
}

int main() {
	pre();
	while (scanf("%s", str) == 1) {
		init();
		n = strlen(str);
		for (int i = 0; i < n; i++) {
			cntv[str[i] - '0']++;
		}
		LL ans = solve(n);
		if (cntv[0]) {
			//扣掉第一位为0的情况
			cntv[0]--;
			ans -= solve(n - 1);
		}
		ans = (ans%mod + mod) % mod;
		printf("%lld
", ans);
	}
	return 0;
}

Notes

确定阶段的划分很重要！比如这题，用不同的数字作为阶段性，非常典型。
往往找到了阶段性，对状态转移方程的设计也会更明确。