一个没有通过,不知道为何
#includeint change(char c) { if(c == 'B') return 0; if(c == 'C') return 1; if(c == 'J') return 2;}int judge(int a, int b) { if(a == b) return 3; else if(a == 'B' && b == 'C') return 1; else if(a == 'B' && b == 'J') return 2; else if(a == 'C' && b == 'B') return 2; else if(a == 'C' && b == 'J') return 1; else if(a == 'J' && b == 'B') return 1; else if(a == 'J' && b == 'C') return 2;}int main() { char str[] = {'B', 'C', 'J'}; int n = 0; scanf("%d", &n); //count[0]:甲赢 count[1]:甲负 count[2]:甲平 //count[3]:乙赢 count[4]:乙负 count[5]:乙平 int count[6] = {0}; //times[0]~times[5]:甲乙分别用锤子，剪刀，布赢的次数 int times[6] = {0}; int id1 = 0, id2 = 0; //甲乙分别赢的最多的手势 for(int i = 0; i < n; i++) { char a, b; getchar(); scanf("%c %c", &a, &b); int temp = judge(a, b); if(temp == 1) { count[0]++; count[4]++; times[change(a)]++;//甲赢使用的手势加一 /*for(int i = 0; i < 6; i++) { printf("--%d", times[i]); } printf("\n"); */ } if(temp == 2) { count[1]++; count[3]++; times[change(b)+3]++;//甲乙赢使用的手势加一 } if(temp == 3) { count[2]++; count[5]++; } } for(int i = 0; i < 3; i++) { if(times[i] > times[id1]) id1 = i; if(times[i+3] > times[id2]) id2 = i; } printf("%d %d %d\n", count[0], count[2], count[1]); printf("%d %d %d\n", count[3], count[5], count[4]); printf("%c %c", str[id1], str[id2]); return 0;}
全部AC
#include#include #include #include #include #include using namespace std;int main(){ int N; scanf("%d",&N); int Jy=0,Jp=0,Js=0; int J[3]={0},Y[3]={0}; //B 0 C 1 J 2 char s1,s2; for(int i=1;i<=N;i++){ getchar(); scanf("%c %c",&s1,&s2); if(s1==s2) Jp++; else if(s1=='C'&&s2=='J'){ Jy++; J[1]++; }else if(s1=='J'&&s2=='B'){ Jy++; J[2]++; }else if(s1=='B'&&s2=='C'){ Jy++; J[0]++; }else{ if(s2=='C') Y[1]++; else if(s2=='J') Y[2]++; else Y[0]++; Js++; } } printf("%d %d %d\n",Jy,Jp,Js); printf("%d %d %d\n",Js,Jp,Jy); if(J[0]>=J[1] && J[0]>=J[2]) printf("B "); else if(J[1]>=J[2]) printf("C "); else printf("J "); if(Y[0]>=Y[1] && Y[0]>=Y[2]) printf("B"); else if(Y[1]>=Y[2]) printf("C"); else printf("J"); printf("\n"); return 0;} --------------------- 作者：fengwuyaQAQ 来源：CSDN 原文：https://blog.csdn.net/fengwuyaQAQ/article/details/85616584 版权声明：本文为博主原创文章，转载请附上博文链接！