POJ 3276 反转问题

#include <cstdio>
#include <iostream> 
#include <cstring>

using namespace std;
int n;
int a[5005][2];
int flag[5005];
int fun(int x){
    int sum  = 0;
    int res = 0;

    memset(flag,0,sizeof(flag));
    for(int i = 0;i+x <= n;i++){
        if(sum%2 == 0 && a[i][0] == ‘B‘ || sum % 2 == 1 && a[i][0] == ‘F‘){
            flag[i] = 1;
            res++;
        }
        else{
            flag[i] = 0;
        }
        sum += flag[i];
        if(i+1-x >= 0){
             sum -= flag[i-x+1];
        }
    }
    for(int i = n-x+1;i < n;i++){
        if(a[i][0] == ‘B‘ && sum % 2 == 0 || sum % 2 == 1&& a[i][0] == ‘F‘){
            return -1;
        }
        if(i-x+1 >= 0){
            sum -= flag[i-x+1];
        }
    }
    return res; 
}
int main(){
    while(cin >> n){
        for(int i = 0;i < n;i++){
            scanf("%s",a[i]);
        }
        int K = 1;
        int M = n;
        for(int i = 1;i <= n;i++){
            int m = fun(i);
            if(m >= 0&&M > m){
                M = m;
                K = i;
            }
        }
        printf("%d %d\n",K,M);
    }
    return 0;
}

反转问题可以用一个标记记录前面反转的次数，如果当前位置前面反转过偶数次那么这个位置相当于没有反转，如果是奇数次说明反转过了！