xiaoxin juju needs help(排列组合+逆元)

描述

传送门：hdu-5651

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.

Examples

• intput

  1 2 3 4

  3 aa aabb a

• output

  1 2 3

  1 2 1

思路

• 其实这道题难点不是推出公式，而是求解公式(吐血.jpg)。
• 设字符串为s,字符串长度为Len,字母i出现次数的二分之一为c[i](取整);
• 如果有解，就要求出现次数是奇数的字母不超过1个(废话)，然后我们只需要考虑一边的所有情况，另一边跟它一样就能构成回文序列。只考虑一边时，共有Len/2个字符,根据高中学的排列组合可知，所有的排列情况是
• 关键就是如何求解这个公式了，由题目可知，Len/2最大是500，也就是要算500!，很明显要爆精度。因为答案对(1e9+7)取摸，这里引入一个转化：
  (a$ \times $b)%m $\rightarrow$ (a%m $ \times $ b%m )%m 显然正确,所以我们求阶乘的时候只要边乘边对(1e9+7)取模就不会爆精度。
  但是 (a $ \div $ b)%m $ \not= $ ((a%m) $ \div $ (b%m))%m,所以我们要用逆元变除法为乘法，就可以边乘边取模。
• 这道题我是用的拓展欧几里德算法求逆元，不知道的可以看我的另一篇博客数论笔记本

代码

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#include #include #include #include #include #include #include using namespace std; #define CRL(a) memset(a,0,sizeof(a)) #define MAX 0xfffffff #define MAXN 6 typedef unsigned long long LL; typedef long long ll; const int mod =1e9+7; ll fac(int n) //求阶乘 { if(n==1||n==0) return 1; else { ll ans=1; for(int i=2;i<=n;i++) ans=(ans*i)%mod; //边乘边取模 return ans; } } int x,y; int exgcd(int a,int b) //拓展欧几里德算法，求出的x,即为a%b下a的逆元 { if(b==0) { x=1;y=0; return a; } int r=exgcd(b,a%b); int c=x; x=y; y=c-a/b*y; return r; } int main() { ll ans; int book[26],n,Flag=0,Len; char s[1005]; cin>>n; while(n--) { CRL(book); Flag=0; ans=0; scanf("%s",s); Len=strlen(s); for(int i=0; i book[s[i]-'a']++; for(int i=0; i<26; i++) if(book[i]&1) Flag++; if(Flag>1) //如果出现次数为奇数的字母超过1,无解 ans=0; else { ans=fac(Len/2); for(int i=0; i<26; i++) if(book[i]>1) { exgcd(fac(book[i]/2),mod); x=x<0? x+mod:x; //x小于0的话要转换成正数 ans=ans*x%mod; } } cout<endl; } return 0; }