HDOJ 5651-xiaoxin juju needs help【字符的全排列】

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1132    Accepted Submission(s): 324

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 integerT(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 mod1,000,000,007.

 

Sample Input

3aaaabba

 

Sample Output

121

 

Source

BestCoder Round #77 (div.2)

 

解题思路：

我们要找到这个字符串可以组成多少个回文串，当然可以看出字母的个数为奇数的个数超过两个包括两个就输出0.

需要推出一个公式p表示字母数量/2.

#include <cstdio>#include <cstring>#define rep(i,j,k) for(i=j;i<k;++i)typedef long long ll;const ll mod = 1000000007;int p[26]; ll a[1001];char str[1005];ll quick_pow(ll a, ll b) {    ll ans = 1;    for (; b; b /= 2, a = (a * a) % mod)        if (b & 1) ans = (ans * a) % mod;    return ans;}int main() {    int t, i, n, odd;    a[0] = 1;    rep(i,1,1001) a[i] = (a[i - 1] * i) % mod;    scanf("%d", &t);    while (t--) {        memset(p, 0, sizeof p);        scanf("%s", str);        n = strlen(str); odd = -1;        rep(i,0,n) ++p[str[i] - 'a'];        ll ans = a[n / 2];        rep(i,0,26) if (p[i] & 1)            if (odd == -1) odd = i;            else { ans = 0; goto end; }        rep(i,0,26) ans = (ans * quick_pow(a[p[i] / 2], mod - 2)) % mod;        end: printf("%lld\n", ans);    }    return 0;}