解题思路:类似于尺取法,依次取 2 0 , 2 1 , 2 2 … … 2^{0},2^{1},2^{2}…… 20,21,22……个数,如果越界就跳出,
#include<bits/stdc++.h>
#define x first
#define y second
#define mem1(h) memset(h,-1,sizeof h)
#define mem0(h) memset(h,0,sizeof h)
#define mcp(a,b) memcpy(a,b,sizeof b)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
namespace IO{
inline LL read(){
LL o=0,f=1;char c=getchar();
while(c<'0'||c>'9'){
if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){
o=o*10+c-'0';c=getchar();}
return o*f;
}
}using namespace IO;
const int N=1e5+7,M=2e5+7,INF=0x3f3f3f3f,mod=1e9+7,P=131;
int a[N];
int n;
int ans;
LL maxv=-1e18;
int main(){
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
for(int d=1,i=1;i<=n;i*=2,d++){
LL sum=0;
for(int j=i;j<i+(1<<d-1)&&j<n;j++){
sum+=a[j];
}
if(sum>maxv){
maxv=sum;
ans=d;
}
}
cout<<ans<<endl;
return 0;
}