HDU 5651 xiaoxin juju needs help

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 665    Accepted Submission(s): 196

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)

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5654 5653 5650 5649 5648 

可以统计每个字符出现的个数，然后分别除以2，也就是对回文串的一半进行排列组合

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <map>

using namespace std;
const __int64 mod=1e9+7;
char a[1005];
int t;
int res[1005];
map<char,int> m;
int a1[1000][1000];
void init()
{
    a1[1][1]=1;
    for(int i=2;i<=600;i++)
    {
        for(int j=1;j<=i;j++)
        {
            if(j==1)
                a1[i][j]=1;
            else if(j==i)
                a1[i][j]=1;
            else
                a1[i][j]=(a1[i-1][j]+a1[i-1][j-1])%mod;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",a);
        int len=strlen(a);
        m.clear();
        int tag=0;
        //cout<<m[a[1]]<<endl;

        for(int i=0;i<len;i++)
        {
            m[a[i]]++;
        }
        int cnt=1;
        init();
        memset(res,0,sizeof(res));
        __int64 sum=0;
        for(int i=0;i<len;i++)
        {
            if(m[a[i]]==-1)
                continue;
             if(m[a[i]]&1)
                tag++;
            res[cnt++]=(m[a[i]])/2;
            sum+=(m[a[i]])/2;
            m[a[i]]=-1;
        }
        if(tag==1&&(!(len&1)))
        {
            printf("0
");
            continue;
        }
        if(tag>=2)
        {
            printf("0
");
            continue;
        }
        __int64 num=1;
        
        for(int i=1;i<cnt;i++)
        {
            num=(num*a1[sum+1][res[i]+1])%mod;
            sum-=res[i];
        }
        printf("%I64d
",num);


    }
    return 0;
}