1 solutions
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0
C :
#include<stdio.h> int main() { int n,i,r,t,x=0,y=0; scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d%d",&r,&t); if(r==t){ continue; }else if(r==1){ if(t==2){ x++; }else{ y++; } }else if(r==2){ if(t==1){ y++; }else{ x++; } }else{ if(t==1){ x++; }else{ y++; } } } if(x==y){ printf("tie"); }else if(x>y){ printf("a win"); }else{ printf("b win"); } return 0; }
C++ :
#include <iostream> using namespace std; int main(){ //a b代表a和b赢的次数 int n,a = 0,b = 0,i,x,y; cin>>n; for(i = 0;i < n;i++){ cin>>x>>y; if(x == 1 && y == 2 || x == 2 && y == 3 || x == 3 && y == 1){ a++; }else if(x != y){ b++; } } if(a > b){ cout<<"a win"<<endl; }else if(a < b){ cout<<"b win"<<endl; }else{ cout<<"tie"<<endl; } }
Python :
n=int(input()) a=[[0 for i in range(2)] for j in range(n)] js1=0 js2=0 for i in range(n): sr=input().split() for j in range(2): a[i][j]=int(sr[j]) if(a[i][0]-a[i][1]==-1 or a[i][0]-a[i][1]==2): js1=js1+1 elif(a[i][0]-a[i][1]==1 or a[i][0]-a[i][1]==-2): js2=js2+1 if(js1>js2): print("a win") elif(js1==js2): print("tie") else: print("b win")
- 1
Information
- ID
- 10402
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
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