p2148 [SDOI2009]E&D

传送门

分析

https://www.luogu.org/blog/flashblog/solution-p2148

代码

#include<bits/stdc++.h>
using namespace std;
int main(){
    int t,n,i,x,y,Ans,cnt;
    scanf("%d",&t);
    while(t--){
      scanf("%d",&n);
      n/=2;
      Ans=0;
      for(i=1;i<=n;i++){
          scanf("%d%d",&x,&y);
          cnt=0;
          x=(x-1)|(y-1);
          while(x&1)cnt++,x>>=1;
          Ans^=cnt;
      }
      if(Ans)puts("YES");
        else puts("NO");
    } 
    return 0;
}

转载于:https://www.cnblogs.com/yzxverygood/p/10357673.html