PAT菜鸡进化史_乙级_1018
大家应该都会玩“锤子剪刀布”的游戏:两人同时给出手势,胜负规则懒得插图了hhh
现给出两人的交锋记录,请统计双方的胜、平、负次数,并且给出双方分别出什么手势的胜算最大。
输入格式:
输入第 1 行给出正整数 N(≤105),即双方交锋的次数。随后 N 行,每行给出一次交锋的信息,即甲、乙双方同时给出的的手势。C
代表“锤子”、J
代表“剪刀”、B
代表“布”,第 1 个字母代表甲方,第 2 个代表乙方,中间有 1 个空格。
输出格式:
输出第 1、2 行分别给出甲、乙的胜、平、负次数,数字间以 1 个空格分隔。第 3 行给出两个字母,分别代表甲、乙获胜次数最多的手势,中间有 1 个空格。如果解不唯一,则输出按字母序最小的解。
输入样例:
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
输出样例:
5 3 2
2 3 5
B B
思路:
这题不难,疯狂复制黏贴改一改就可以了hhh
Code:
#include <iostream>
int main(){
using namespace std;
// input the game times
int n;
cin >> n;
// count the result
int win_jia[3] = {0}, win_yi[3] = {0}; // B_win C_win J_win
int tie = 0; // tie
char jia, yi;
for (int i = 0; i < n; i++){
cin >> jia >> yi;
if (jia == 'B' && yi == 'C')
win_jia[0]++;
else if (jia == 'C' && yi == 'J')
win_jia[1]++;
else if (jia == 'J' && yi == 'B')
win_jia[2]++;
else if (jia == 'C' && yi == 'B')
win_yi[0]++;
else if (jia == 'J' && yi == 'C')
win_yi[1]++;
else if (jia == 'B' && yi == 'J')
win_yi[2]++;
}
int win_jia_num = win_jia[0] + win_jia[1] + win_jia[2];
int win_yi_num = win_yi[0] + win_yi[1] + win_yi[2];
// output the result
cout << win_jia_num << " " << n - win_jia_num - win_yi_num << " " << win_yi_num << endl;
cout << win_yi_num << " " << n - win_jia_num - win_yi_num << " " << win_jia_num << endl;
char temp[3] = {'B', 'C', 'J'};
int max_jia = win_jia[0] >= win_jia[1] ? 0 : 1;
max_jia = win_jia[max_jia] >= win_jia[2] ? max_jia : 2;
int max_yi = win_yi[0] >= win_yi[1] ? 0 : 1;
max_yi = win_yi[max_jia] >= win_yi[2] ? max_yi : 2;
cout << temp[max_jia] << " " << temp[max_yi] << endl;
return 0;
}