PAT 1040 有几个PAT (25 分)

#include <iostream>
#include <string>
using namespace std;
int main() {
    string s;
    cin >> s;
    int len = s.length(), result = 0, count_p = 0, count_t = 0;
    for (int i = 0; i < len; i++) {
        if (s[i] == 'T')
            count_t++;
    }
    for (int i = 0; i < len; i++) {
        if (s[i] == 'P') count_p++;
        if (s[i] == 'T') count_t--;
        if (s[i] == 'A') 
        result = (result + (count_p * count_t) % 1000000007) % 1000000007;
    }
    cout << result;
    return 0;
}

方法2

#include <stdio.h>
#define LIM 1000000007
int main()
{
    int P = 0, PA = 0, PAT = 0;
    char c;
    while((c = getchar()) != '\n')
    {
        if(c == 'P')   P++;
        if(c == 'A')   PA = (PA + P) % LIM;
        if(c == 'T')   PAT = (PAT + PA) % LIM;
    }
    printf("%d", PAT);
    return 0;
}
©著作权归作者所有,转载或内容合作请联系作者
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

推荐阅读更多精彩内容