[2016-03-26][codeforces][632][B][Alice, Bob, Two Teams]

• 时间:2016-03-26 10:14:35 星期六

• 题目编号:[2016-03-26][codeforces][632][B][Alice, Bob, Two Teams]

• 题目大意:给定n段字符串,每段字符串要么都是A,要么都是B,已知每段字符串的长度和字符,问把前k段字符或者后k段字符翻转(A -> B,b -> A,k是任意的),问最多能得到多少个B

• 分析:先求前n段字符总字符数的前缀和,和B字符数的前缀和,枚举翻转的长度,更新最大值

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 5 * 1E5 + 10;
LL a[maxn],b[maxn];
char str[maxn];
int main(){
        int n;       
        scanf("%d",&n);
        for(int i = 1;i <= n ;++i){
                scanf("%I64d",&a[i]);               
        }
        getchar();
        gets(str+1);
        for(int i = 1 ;i <= n ;++i){               
                if(str[i] == 'B'){
                        b[i] = b[i - 1] + a[i];
                }else b[i] = b[i - 1];
                a[i] += a[i - 1];
        }
        LL ans = 0;
        for(int i = 0;i <= n ;++i){
                ans = max(ans, a[i] - b[i] + (b[n] - b[i]));
                ans = max(ans,(a[n] - a[i]) - (b[n] - b[i]) + b[i]);
        }
        printf("%I64d\n",ans);
        return 0;
}

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