HDU 5651xiaoxin juju needs help

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1310    Accepted Submission(s): 378


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(T20) which represents the number of test cases.
For each test case, there is a single line containing a string S(1length(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
 
排列组合问题。C[n][m]用dp求一下。
然后对于aabababbcc这组样例  我们取1/2进行考虑,aabbc.    答案就是c(5,2)*c(3,2)*c(1,1)
 
 
/* ***********************************************
Author        :
Created Time  :2016/3/30 15:28:22
File Name     :hdu5651.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 1000000007 
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
char s[maxn];
int cnt[30];
int c[510][510];
void init(){
    memset(c,0,sizeof c);
    for(int i=0;i<510;i++){
        c[i][1]=i;c[i][0]=1;
    }
    for(int i=2;i<=500;i++){
        for(int j=2;j<=i;j++){
            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
        }
    }
}
void solve(int n){
    n/=2;
    ll ans=1,sum=n;
    for(int i=0;i<26;i++){
        ans*=c[n][cnt[i]/2];
        ans%=mod;
        n-=cnt[i]/2;
    }
    cout<<ans<<endl;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    init();
    while(t--){
        scanf("%s",&s);
        cle(cnt);
        int n=strlen(s);
        for(int i=0;i<n;i++){
            cnt[int(s[i]-'a')]++;
        }
        int mark=0;
        for(int i=0;i<26;i++){
            if(cnt[i]&1){
                mark++;
            }
        }
        if(n%2==0){
            if(mark>0){
                printf("%d
",0);continue;
            }
            solve(n);
        }
        else{
            if(mark==1){
                solve(n);
            }
            else{
                printf("%d
",0);
            }
        }

    }
    return 0;
}
原文地址:https://www.cnblogs.com/pk28/p/5338782.html