HDU5651 xiaoxin juju needs help 组合

xiaoxin juju needs help

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

Total Submission(s): 293 Accepted Submission(s): 87



[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3
aa
aabb
a

[align=left]Sample Output[/align]

1
2
1

[align=left]Source[/align]
BestCoder Round #77 (div.2)

先看答案为0的情况，如果有大于1种的字符的个数是奇数，那么答案为0。

如果就一种字符答案为1.

形成回文，只需取每种字符的一半进行排列，后面的一半补上去就能形成回文了，故重点就是求排列的个数。

排列的个数其实就是有重复元素的全排列的个数= tot!/(a!*b!*...*z!)，其中a,b,..,z代表对应字符的个数的一半，tot也是总和的一半。

过程需要求一下逆元。

/****************
*PID:hdu5651
*Auth:Jonariguez
*****************
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef long long Ll;
const int maxn=1000+10;
const int MOD=1e9+7;
Ll quick_pow(Ll a,Ll b){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

char str[maxn];
int vis[30];
LL fact[maxn];

int main()
{
int i,j,n,m,T;
fact[0]=1;
for(i=1;i<=1001;i++)
fact[i]=(fact[i-1]*i)%MOD;  //预处理阶乘
scanf("%d",&T);
while(T--){
scanf("%s",str+1);
n=strlen(str+1);
int f=0;
memset(vis,0,sizeof(vis));
for(i=1;i<=n;i++)
vis[str[i]-'a']++;
int cnt1=0;
for(i=0;i<26;i++){
if(vis[i]) cnt1++;
if(vis[i]&1){
if(f) break;
f=i+1;
}
}
if(i<26){
puts("0");
continue;
}
if(cnt1==1){
puts("1");continue;
}
LL res=1;
int cnt=0;
for(i=0;i<26;i++){
if(vis[i]){
cnt+=(vis[i]/2);
}
}
res=fact[cnt];
for(i=0;i<26;i++){
if(vis[i]){
LL inv=quick_pow(fact[vis[i]/2],MOD-2); //由费马小定理求出逆元
res=(res*inv)%MOD;
}
}
printf("%I64d\n",res);
}
return 0;
}