noi.ac #227 random

分析

我们发现实际只要计算a[i]>b[j]和a[i]<b[j]哪种多即可

代码

#include<bits/stdc++.h>
using namespace std;
int a[],b[],n,Ans,le,ri;
int main(){
    int i,j,k,t;
    scanf("%d",&t);
    while(t--){
      Ans=,le=;
      scanf("%d",&n);
      for(i=;i<=n;i++)scanf("%d",&a[i]);
      for(i=;i<=n;i++)scanf("%d",&b[i]);
      sort(a+,a+n+);
      sort(b+,b+n+);
      for(i=;i<=n;i++){
          while(le<=n&&b[le+]<a[i])le++;
          ri=le;
          while(ri<=n&&b[ri+]==a[i])ri++;
          Ans+=le+ri-n;
      }
      if(Ans>)puts("Alice");
        else if(Ans<)puts("Bob");
        else puts("Tie");
    }
    return ;
}