PAT B1018

PAT B1018"/>

PAT B1018

牛客网PAT乙级第八题，锤子剪刀布，题目描述大致如下：

输入第一行正整数N(N<=105)，代表交锋次数，接下来N行输入交锋信息，格式"甲 乙"，C代表锤子，B代表布，J代表剪刀。

输出第1，2行分别给出甲和乙的胜，负，平的次数，数字间空格分离，第三行给出两个字母，代表甲和乙获胜次数最多的手势。

空格分隔，如果解不唯一，输出字母序最小的解。

代码如下：

#include <iostream>using namespace std;char FindVictoryWay(int shitou, int jiandao, int bu)
{if (shitou > jiandao){if (bu >= shitou){return 'B';}else{return 'C';}}else if (shitou == jiandao){if (shitou > bu){return 'C';}else{return 'B';}}else{if (jiandao > bu){return 'J';}else{return 'B';}}}int main()
{int N;//交锋次数char a_result, b_result;int a_victory = 0, a_ping = 0, b_victory = 0;int a_shitou = 0, a_jiandao = 0, a_bu = 0, b_shitou = 0, b_jiandao = 0, b_bu = 0;char a_winmost, b_winmost;while (cin >> N){for (int i = 0; i < N; i++){cin >> a_result >> b_result;if (a_result == b_result){a_ping++;}else if (((a_result == 'C') && (b_result == 'J')) || ((a_result == 'B') && (b_result = 'C')) || ((a_result == 'J') && (b_result == 'B'))){a_victory++;switch (a_result){case 'C':a_shitou++;break;case 'J':a_jiandao++;break;case 'B':a_bu++;break;default:break;}}else{b_victory++;switch (b_result){case 'C':b_shitou++;break;case 'J':b_jiandao++;break;case 'B':b_bu++;break;default:break;}}}a_winmost = FindVictoryWay(a_shitou, a_jiandao, a_bu);b_winmost = FindVictoryWay(b_shitou, b_jiandao, b_bu);cout << a_victory << " " << a_ping << " " << b_victory << endl;cout << b_victory << " " << a_ping << " " << a_victory << endl;cout << a_winmost << " " << b_winmost << endl;}return 0;
}