杭电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): 1358    Accepted Submission(s): 387


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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挺好的一道题，来讲个规律吧，(a+b+c+d+...z)的阶乘==c(a+b+c....+y+z,a+b+c+d+...y)*c(a+b+c+d+......+x+y,a+b+c+d+....+x),*....c(a+b,a);

根据(a+b)！=c(a+b,a)推出来的，那么题目可以转换成，已知有m个字母，每个字母有不同的个数，问排列起来有多少种排列方式，就用到了上边的公式，因为题目所用的数据比较大，所以求组合数的函数不能用，需要用到杨辉三角，杨辉三角和组合数的过程是一样的，那么便得到了答案：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
char s[110000];
char c[11000];
__int64 i,j,k,l,m,n,p,biaoji[1100],ch[2010][2010];
//__int64 ac(__int64 m,__int64 n)//组合函数不对，因为当ans大到一定程度的时候
//{//有一个取余的过程，这个过程可能造成ans/i不为整数
// __int64 ans=1;
// __int64 k=m;
// //n=min(n,m-n);
// for(__int64 i=1;i<=n;i++)
// {
// ans*=(k-i+1);
// ans/=i;
// ans%=1000000007;
// }
// return ans;
//}
int main()
{
memset(ch,0,sizeof(ch));
ch[0][0]=1;//需要用杨辉三角来代替求组合数的函数
for( i=1; i<=510; i++)
{
ch[i][0]=1;
for( j=1; j<=510; j++)
{
ch[i][j]=(ch[i-1][j]+ch[i-1][j-1])%1000000007;
}
}
scanf("%I64d",&p);
while(p--)
{
scanf("%s",s);
map<char , int> q;//map求出每个字母对应的个数
l=strlen(s);
q.clear();
int cnt=0;
for(i=0;i<l;i++)
{
if(!q[s[i]])
{
c[cnt]=s[i];
cnt++;
q[s[i]]++;
}
else
q[s[i]]++;
}
int cmt=0;
for(i=0;i<cnt;i++)
{
if(q[c[i]]%2)
{
cmt++;
q[c[i]]--;
}
}
if(cmt>1)
printf("0\n");
else
{
int nn=q[c[0]]/2;
__int64 ans=1;
for(i=1;i<cnt;i++)
{
//ans*=ac(nn+q[c[i]]/2,nn);
ans*=ch[nn+q[c[i]]/2][nn];
ans%=1000000007;
nn+=q[c[i]]/2;
//printf("%I64d\n",ac(nn+q[c[i]]/2,nn));
}
printf("%I64d\n",ans);
}
// printf("%I64d\n",ac(44,42));
// printf("%I64d\n",ac(44,2));
// printf("%I64d\n",ac(42,21));
}
}