生活大爆炸版石头剪刀布

传送门：https://www.luogu.org/problemnew/show/P1328#sub

这道题相信大家都会做（

除了下面贴的代码外，还有个比较聪明的办法就是手动打出vs[i][j]表示加分情况

#include<cstdio>
using namespace std;
int n,na,nb,suma,sumb;
int a[205],b[205];
int main()
{
    scanf("%d%d%d",&n,&na,&nb);
    for(int i = 0;i < na;i++) scanf("%d",&a[i]);
    for(int i = 0;i < nb;i++) scanf("%d",&b[i]);
    for(int i = 0;i < n;i++)
    {
        if(a[i%na] == 0 && b[i%nb] == 1) sumb++;
        else if(a[i%na] == 0 && b[i%nb] == 2) suma++;
        else if(a[i%na] == 0 && b[i%nb] == 3) suma++;
        else if(a[i%na] == 0 && b[i%nb] == 4) sumb++;
        else if(a[i%na] == 1 && b[i%nb] == 2) sumb++;
        else if(a[i%na] == 1 && b[i%nb] == 3) suma++;
        else if(a[i%na] == 1 && b[i%nb] == 4) sumb++;
        else if(a[i%na] == 2 && b[i%nb] == 3) sumb++;
        else if(a[i%na] == 2 && b[i%nb] == 4) suma++;
        else if(a[i%na] == 3 && b[i%nb] == 4) suma++;
        
        else if(a[i%na] == 1 && b[i%nb] == 0) suma++;
        else if(a[i%na] == 2 && b[i%nb] == 0) sumb++;
        else if(a[i%na] == 3 && b[i%nb] == 0) sumb++;
        else if(a[i%na] == 4 && b[i%nb] == 0) suma++;
        else if(a[i%na] == 2 && b[i%nb] == 1) suma++;
        else if(a[i%na] == 3 && b[i%nb] == 1) sumb++;
        else if(a[i%na] == 4 && b[i%nb] == 1) suma++;
        else if(a[i%na] == 3 && b[i%nb] == 2) suma++;
        else if(a[i%na] == 4 && b[i%nb] == 2) sumb++;
        else if(a[i%na] == 4 && b[i%nb] == 3) sumb++;
    }
    printf("%d %d",suma,sumb);
    return 0;
 }