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Codeforces Round #803 (Div. 2)
2022-07-06 10:01:00 【Stellaris_L】
A - XOR Mixup
题目大意:
给一个有 n 个数的数组,其中有 1 个数是另外 n-1 个数的异或和,求出这个数。
做法:
因为必然有解,所以整个数组的异或和为 0,只需要随意选择一个就可以了。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10;
int T,n;
void slove(){
int n;
cin>>n;
int ans;
for(int i=0;i<n;i++){
cin>>ans;
}
cout<<ans<<endl;
}
int main(){
cin>>T;while(T--)
slove();
return 0;
}
B - Rising Sand
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