hdu5651 xiaoxin juju needs help 组合数

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 594    Accepted Submission(s): 175

Problem Description

As we all known, xiaoxin is a brilliant coder. He knew **palindromic** strings when he was only a six grade student at elementry school.

This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin's leader needs to buy?

 

Input

This problem has multi test cases. First line contains a single integer  T(T≤20)  which represents the number of test cases.
For each test case, there is a single line containing a string  S(1≤length(S)≤1,000) .

 

Output

For each test case, print an integer which is the number of watermelon candies xiaoxin's leader needs to buy after mod  1,000,000,007 .

 

Sample Input

   
   
   
   

    
    
    
    3
aa
aabb
a
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
1
   
   
   
   

 

Source

BestCoder Round #77 (div.2)

无语了，求杨辉三角居然忘了注意边界C(n, 0) = 1，求的时候取模还不彻底，无语......失之毫厘谬以千里......

思路主要就是判断字符串长度的奇偶性，结合每个元素的奇偶性来判断有没有可能组成回文

如果可以的话然后利用组合数来算有多少种

比如

cccbba a abbccc

len = 13 但是实际上只要算len / 2 = 6也就是黑体的这一半的组合情况就行了，另外一半对称

算法是，如果取a,b,c的顺序，那么对于a:C(6, 1)，对于b去掉a占的位置:C(6 - 1, 2) = C(5, 2)

对于c去掉b, a占得位置:C(6 - 1 - 2, 3) = C(3, 3)

根据乘法原理ans = C(6, 1) * C(5, 2) * C(3, 3)

偶数情况以此类推

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>

using namespace std;

long long mod = 1000000007;

char ch[1005];
long long c[1005][1005];
map<char, int> m;

//用记忆化搜索来求杨辉三角，记忆数组c
long long C(int n, int m) {
	if (n == m || m == 0) return 1;  //注意边界
	if (c[n][m] != -1) {
		return c[n][m] % mod; //注意取模
	}
	return c[n][m] = (C(n - 1, m) + C(n - 1, m - 1)) % mod;  //注意取模
}

int main()
{
	int T;
	//c是通用的直接放最外面就行
	memset(c, -1, sizeof(c));
	c[0][0] = 1;
	c[1][0] = 1; c[1][1] = 1;
	c[2][0] = 1; c[2][1] = 2; c[2][2] = 1;
	scanf("%d", &T);
	while (T--) {
		m.clear();
		scanf("%s", ch);
		int len = strlen(ch);
		for (int i = 0; i < len; i++) {
			m[ch[i]]++;  //散列计数，其实直接开个数组也行
		}
		if (len == 1) {
			puts("1");
			continue;
		}

		//算奇数的个数
		int ji = 0;
		for (map<char, int>::iterator i = m.begin(); i != m.end(); i++) {
			if (i->second % 2) {
				ji++;
			}
		}

		long long ans = 1;
		//能算的情况是没有奇数，而且字符串的长度为偶数，或者有且仅有一个奇数字符串长度也为奇数
		//否则不可能构成回文
		if ((ji == 0 && len % 2 == 0) || (ji == 1 && len % 2 != 0)) {
			//只需要计算一般的情况就可以了，因为另外一般是对称的
			len /= 2;
			for (map<char, int>::iterator i = m.begin(); i != m.end(); i++) {
				i->second /= 2;
			}
			for (map<char, int>::iterator i = m.begin(); i != m.end(); i++) {
				ans = (ans * (C(len, i->second) % mod)) % mod; //注意取模
				len -= i->second;
			}
		}
		else {
			puts("0");
			continue;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}