HDU 5651 xiaoxin juju needs help（组合数学）

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1657    Accepted Submission(s): 499

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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刚开始看到这道题时下了一跳，后来仔细想想属于组合数学的知识，题意可以转换为有几种不同的字母，每种字母都有不同的数量，问有多少种不同的排列方式，不过有点无奈的是以前学的组合数学知识差不多忘完了。。。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[510][510];
void zuo()
{
	int i,j;
	for(i=0;i<=500;i++)
	{
		a[i][0]=1;
	}
    for(i=0;i<=500;i++)
	for(j=1;j<=i;j++)
	{
		a[i][j]=(a[i-1][j-1]+a[i-1][j])%1000000007;
	}
}
int main()
{
	int t,i;
	char s[110000];
	int num[30];
	zuo();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",s);
		memset(num,0,sizeof(num));
		int len=strlen(s);
		for(i=0;i<len;i++)
		{
			num[s[i]-'a'+1]++;
		}
		int odd=0;
		for(i=1;i<=26;i++)
		{
			if(num[i]%2!=0)
			{
				odd++;
				num[i]=0;
			}
			else
			num[i]=num[i]/2;
		}
		if(odd>1||(odd==1&&len%2==0))
		{
			printf("0\n");
			continue;
		}
		__int64 ans=1;
		len=len/2;
		for(i=1;i<=26;i++)
		{
			ans=(ans*a[len][num[i]])%1000000007;
			len=len-num[i];
		}
		printf("%I64d\n",ans);
	}
	return 0;
}