PAT basic 1018

来源:互联网 发布:烈焰传奇翅膀进阶数据 编辑:程序博客网 时间:2024/05/29 17:52
#include <iostream>using namespace std;int main() {    int n;    cin >> n;    int jiawin = 0, yiwin = 0;    int jia[3] = {0}, yi[3] = {0};    for (int i = 0; i < n; i++) {        char s, t;        cin >> s >> t;        if (s == 'B' && t == 'C') {            jiawin++;            jia[0]++;        } else if (s == 'B' && t == 'J') {            yiwin++;            yi[2]++;        } else if (s == 'C' && t == 'B') {            yiwin++;            yi[0]++;        } else if (s == 'C' && t == 'J') {            jiawin++;            jia[1]++;        } else if (s == 'J' && t == 'B') {            jiawin++;            jia[2]++;        } else if (s == 'J' && t == 'C') {            yiwin++;            yi[1]++;        }    }    cout << jiawin << " " << n - jiawin - yiwin << " " << yiwin << endl << yiwin << " " << n - jiawin - yiwin << " " << jiawin << endl;    int maxjia = jia[0] >= jia[1] ? 0 : 1;    maxjia = jia[maxjia] >= jia[2] ? maxjia : 2;    int maxyi = yi[0] >= yi[1] ? 0 : 1;    maxyi = yi[maxyi] >= yi[2] ? maxyi : 2;    char str[4] = {"BCJ"};    cout << str[maxjia] << " " << str[maxyi];    return 0;}
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