1018. 锤子剪刀布 (20)

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#include<iostream>#include<math.h>#include<string>using namespace std;int cmp(string a, string b,int& winB,int& winC,int& winJ){if (a == b)return 0;if (a == "C"){if (b == "B")return -1;if (b == "J"){winC++;return 1;}}if (a == "B"){if (b == "J")return -1;if (b == "C"){winB++;return 1;}}if (a == "J"){if (b == "C")return -1;else if(b=="B"){winJ++;return 1;}}}char com(int b, int c, int j){if (b > c){if (b >=j)return 'B';if (b < j)return 'J';}if (b == c){if (b >= j)return 'B';elsereturn 'J';}if (b < c){if (c >= j)return 'C';if (c < j)return 'J';}}int main(){int n;cin >> n;string jia, yi;int aWin = 0, aLose = 0, aMean = 0, bWin = 0, bLose = 0, bMean = 0;int aWinB=0, aWinC = 0, aWinJ = 0, bWinB = 0, bWinC = 0, bWinJ = 0;while (n--){cin >> jia >> yi;int countA = cmp(jia,yi,aWinB,aWinC,aWinJ);cmp(yi, jia, bWinB, bWinC, bWinJ);if (countA == -1){aLose++;bWin++;}if (countA == 0){aMean++;bMean++;}if (countA == 1){aWin++;bLose++;}}cout << aWin << ' ' << aMean << ' ' << aLose << endl;cout<< bWin << ' ' << bMean << ' ' << bLose << endl;cout << com(aWinB, aWinC, aWinJ) << ' ' << com(bWinB, bWinC, bWinJ);return 0;}

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