【HDOJ 5652】xiaoxin juju needs help（排列组合）

【HDOJ 5652】xiaoxin juju needs help（排列组合）

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1122    Accepted Submission(s): 322

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)

 

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题目大意:T组输入，每组一个字符串。只由小写字母组成。

要求通过交换该串中某些字符的位置，让它变成一个回文串。问可以得到多少种不同的回文串。

题目拐了个弯。其实就是问给一些字符，用它们能组成多少种不同的字符串……

首先可以知道，如果每种字符都是偶数个，一定可以组成回文串（对称放）

如果有奇数个，奇数个的字符只能有一种，否则无法构成回文串(奇数的字符放到中心，这样剩下的由全变成偶数了，关于这个奇数字符中心对称就是。

接下来就是对于能组成回文串的字符进行放置。

我的方法是用组合数。

只放一半的字符，另一半对称即可。

如果总长度len 那么需要放置的就是len/2个字符。

这样枚举每种字符 假设m为放置的字符个数(len/2) vi表示第i种字符数量的一半(如上，只放一半，另一半对称)

结果就是C(m,v1)*C(m-v1,v2)*C(m-v1-v2,v3)*...*C(v26,v26)

递推搞一下就出来了。组合数可以递归预处理出来。

另一种直接用定理，。

如果每个字符都当做不一样的话，总的组合就是n!(n为字符个数

然后出去重复，就是n!/v1!/v2!/v3!/..../v26!

除阶乘没法取余，会炸，反过来求逆元即可，同样可以预处理出来。

逆元的没写，上下组合的吧：

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

int cnt[26];

LL C[555][555];

LL cal(int m,int n)
{
	if(~C[m][n]) return C[m][n];
	if(m == n) return C[m][n] = 1;
	if(n == 0) return C[m][n] = 1;
	if(m == 0) return C[m][n] = 0;
	return C[m][n] = (cal(m-1,n)+cal(m-1,n-1))%mod;
}

int main()
{
	//fread();
	//fwrite();

	memset(C,-1,sizeof(C));

	for(int i = 500; i >= 0; --i)
		for(int j = 500; j >= 0; --j)
			if(C[i][j] == -1) cal(i,j);

	int t;
	char ch;
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		memset(cnt,0,sizeof(cnt));
		while((ch = getchar()) != '\n' && ch != '\r') cnt[ch-'a']++;
		int c = 0,len = 0;
		for(int i = 0; i < 26; ++i)
		{
			if(cnt[i]&1) c++;
			len += cnt[i];
		}
		
		if(c > 1)
		{
			puts("0");
			continue;
		}

		LL ans = 1;
		len >>= 1;
		for(int i = 0; i < 26; ++i)
		{
			ans = (ans*C[len][cnt[i]>>1])%mod;
			len -= cnt[i]>>1;
		}
		printf("%lld\n",ans);
	}

	return 0;
}