题解

跳出条件i*2<=2时

#include<stdio.h>
long long n;
int times;
long long i=1;
int main()
{
    scanf("%lld",&n);
    for(i=1;i*2<=n;i*=2)
        {
            times++;
        }
    printf("%d\n",times);
    for(i=1;i*2<=n;i*=2)
        {
            printf("%lld\n",i);
        }
}