analysis
First split the binary and count 1 number
because there can be no two 0s or two 1s. Count the number of 1s in the binary of each number. If there is only 1 digit 1, then directly count 2 i 2^i2i
or the rest can be 1, you can choose to add up to2 j 2^j2j can
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//先 拆分二进制 统计1个数
//然后 10^9大概是30位 for(i=30~0) 如果1的个数=1 就累加2的i次方 >1就
//for(j=i~0) 累加2^j
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,a[105],cnt[105],ans;
void count_one(int x)
{
int j=0;
while(x)
{
if(x%2==1) cnt[j]++;
x/=2;
j++;
}
}
int ksm(int a,int n)
{
long long ans=1;
while(n)
{
if(n&1)
{
ans=(ans*a);
}
a=a*a;
n>>=1;
}
return ans;
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
count_one(a[i]);
}
for(register int i=30;i>=0;i--)
{
if(cnt[i]==1) ans+=ksm(2,i);
if(cnt[i]>=2)
{
for(register int j=i;j>=0;j--)
{
ans+=ksm(2,j);
}
break;
}
}
cout<<ans;
return 0;
}