与众不同的做法
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与众不同的做法
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bajia楼主2024/10/26 21:07
#include<bits/stdc++.h>
using namespace std;
bool a[5][15];
int a1[128],a2[128],ans=52;
int n;
char c1,c2;
int main(){
	a1['D']=0;
	a1['C']=1;
	a1['H']=2;
	a1['S']=3;
	a2['A']=0;
	a2['2']=1;
	a2['3']=2;
	a2['4']=3;
	a2['5']=4;
	a2['6']=5;
	a2['7']=6;
	a2['8']=7;
	a2['9']=8;
	a2['T']=9;
	a2['J']=10;
	a2['Q']=11;
	a2['K']=12;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>c1>>c2;
		if(a[a1[c1]][a2[c2]]==0){
			a[a1[c1]][a2[c2]]=1;
			ans--;
		}
	}
	cout<<ans;
	return 0;
}

2024/10/26 21:07
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