hdu3363 Ice-sugar Gourd(杂题）

枚举n/2个区间，判断 区间内的h和t是不是总数一半，是的话区间左边切一刀，右边切一刀（ 一共最多切两刀）

 

题目：

Ice-sugar Gourd

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 740    Accepted Submission(s): 264

Problem Description

Ice-sugar gourd, “bing tang hu lu”, is a popular snack in Beijing of China. It is made of some fruits threaded by a stick. The complicated feeling will be like a both sour and sweet ice when you taste it. You are making your mouth water, aren’t you?

I have made a huge ice-sugar gourd by two kinds of fruit, hawthorn and tangerine, in no particular order. Since I want to share it with two of my friends, Felicia and his girl friend, I need to get an equal cut of the hawthorns and tangerines. How many times will I have to cut the stick so that each of my friends gets half the hawthorns and half the tangerines? Please notice that you can only cut the stick between two adjacent fruits, that you cannot cut a fruit in half as this fruit would be no good to eat.

 

 

Input

The input consists of multiply test cases. The first line of each test case contains an integer, n(1 <= n <= 100000), indicating the number of the fruits on the stick. The next line consists of a string with length n, which contains only ‘H’ (means hawthorn) and ‘T’ (means tangerine).
The last test case is followed by a single line containing one zero.

 

 

Output

Output the minimum number of times that you need to cut the stick or “-1” if you cannot get an equal cut. If there is a solution, please output that cuts on the next line, separated by one space. If you cut the stick after the i-th (indexed from 1) fruit, then you should output number i to indicate this cut. If there are more than one solution, please take the minimum number of the leftist cut. If there is still a tie, then take the second, and so on.

 

 

Sample Input

4 HHTT 4 HTHT 4 HHHT 0

 

 

Sample Output

2 1 3 1 2 -1

 

 

Source

“光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛

 

 

Recommend

lcy

 

 

代码：

 

//Sat Jan 11 13:25:58 2014
//Author:Minshik
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <memory.h>
#include <cctype>
#include <cmath>
#include <list>
using namespace std;

char sw[100000+10];
int pH[100000+10];
int pT[100000+10];
int main()
{
    freopen("IN","r",stdin);
    std::ios::sync_with_stdio(false);
    int n;
    while(cin>>n && n!=0)
    {
        int sh=0,st=0;
        int ca=0,cb=0;
        cin.ignore();
        for(int i=0;i<n;i++)
        {
            cin>>sw[i];
            if(sw[i]=='H')sh++;
        }
        st= n-sh;

        //cout<<sw<<endl;
        if(st%2==1 || sh%2==1)cout<< "-1"<<endl;
        else
        {
            int ppt=0,pph=0;

                for(int j=0;j<n/2;j++)
                {
                    if(sw[j]=='H')pph++;
                    if(sw[j]=='T')ppt++;

                }
            int flag =false;
                if(ppt==st/2 && pph==sh/2){cout<<'1'<<endl<<n/2<<endl;flag=true;}

            if(!flag)
            for(int i=1;i+n/2<n;i++)
            {
                if(sw[i-1]=='H'){pph--;}
                else ppt--;

                if(sw[i+n/2-1]=='H')pph++;
                else ppt++;

                //cout<<"okok"<<i<<' '<<i+n/2-1<<' '<<pph<<' '<<ppt<<endl;
                if(ppt==st/2 && pph==sh/2){cout<<'2'<<endl<<i<<' '<<i+n/2<<endl;flag=true;break;}
            }

            if(!flag)
                cout<<'-1'<<endl;

        }

    }


    return 0;
}