题解 | #Branch of Faith#
Branch of Faith
https://ac.nowcoder.com/acm/contest/120563/J
题目大概意思就是1到n的节点依次排列成二叉树,根据题目画图,我是根据给的询问节点编号依次除二得到深度,最后特判一下可能没填满的情况
#include <bits/stdc++.h>
using namespace std;
int main(){
long long t;
cin>>t;
while(t--){
long long n,q;
cin>>n>>q;
for(long long i=1;i<=q;i++){
long long x,sum=1;
cin>>x;
while(x!=1){
x/=2;
sum++;
}
long long max=pow(2,sum)-1;
long long min=pow(2,sum-1);
if(n<max)
cout<<n-min+1<<"\n";
else
cout<<min<<"\n";
}
}
}
using namespace std;
int main(){
long long t;
cin>>t;
while(t--){
long long n,q;
cin>>n>>q;
for(long long i=1;i<=q;i++){
long long x,sum=1;
cin>>x;
while(x!=1){
x/=2;
sum++;
}
long long max=pow(2,sum)-1;
long long min=pow(2,sum-1);
if(n<max)
cout<<n-min+1<<"\n";
else
cout<<min<<"\n";
}
}
}

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