poj--5651

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1164    Accepted Submission(s): 335

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

3aaaabba

 

Sample Output

121

 

Source

BestCoder Round #77 (div.2)

题目大意:当字符串中含有的字符,其个数为奇数的字符种类>1时,输出0;剩下的就可以理解成给你一串含有重复字母的字符串,问有多少种组合方式.

用到公式n!/(pa!*pb!*pc!......pz!),其中在这题中n=l/2,p？=？字符的总数除以2。因为数很大，所以可以边乘边取模，但是分子不能直接除以分母，需要用到逆元的知识。

代码如下：

#include<cstdio>#include<string.h>#define mod 1000000007char s[1010];int num[27];long long power[600];void inin(){int i;power[0]=1;for(i=1;i<=500;i++){power[i]=power[i-1]*i%mod;}} void gcd(long long a, long long  b,  long  long &x, long long  &y) {    if (!b) {                x = 1; y = 0;    }    else {        gcd(b, a%b, y, x);//y=x',x=y';        //x=y'; y=x'-(a/b)*y';        y -= (a / b)*x;    }}int main(){int t,i,j,l,v;long long ans;long long sum;inin();scanf("%d",&t);while(t--){scanf("%s",s);l=strlen(s);memset(num,0,sizeof(num));for(i=0;i<l;i++){num[s[i]-'a']++;}v=0;sum=1;for(i=0;i<26;i++){if(num[i]){if(num[i]%2)v++;sum=sum*power[num[i]/2]%mod;}if(v>=2){break;}}if(v>=2)printf("0\n");else {long long  x,y;//printf("%lld\n",ans/sum%mod);逆元定理 gcd(sum,mod,x,y);x = (x%mod + mod) % mod;            long long  ans = (power[l / 2] * x) % mod;            printf("%lld\n", ans);}}return 0;}