HDU 3305 Ice-sugar Gourd

Ice-sugar Gourd

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 936    Accepted Submission(s): 329

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

 

 

开始用双重for循环，超时了，还是要用一个数组nh[i]来记录到第i个有多少个h

然后从前往后的结果

 

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 123456
int n,cnt_h,cnt_t;
char a[N];
int nh[N];

int main()
{
    while(~scanf("%d",&n))
    {
        getchar();
        if(n==)break;

        cnt_h=;
        for(int i=;i<n;i++)
        {
            scanf("%c",&a[i]);
            if(a[i]=='H')cnt_h++;
            nh[i]=cnt_h;
        }
        if(n&)
        {
            cout<<-<<endl;
            continue;
        }
        if(cnt_h&)
        {
            cout<<-<<endl;
            continue;
        }
        if(nh[n/-]==cnt_h/)
        {
            printf("1\n%d\n",n/);
            continue;
        }
        for(int i=;i<n;i++)
        {
            if(nh[n/+i-]-nh[i-]==cnt_h/)
            {
                printf("2\n%d %d\n",i,i+n/);
            break;
            }

        }
    }
    return ;
}

//freopen("1.txt", "r", stdin);
    //freopen("2.txt", "w", stdout);
//**************************************