3 solutions
- 1
Information
- ID
- 949
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 3
- Tags
- (None)
- # Submissions
- 74
- Accepted
- 23
- Uploaded By
#include<bits/stdc++.h>
using namespace std;
const int N=20;
int a[N];
int main()
{
int n;
cin>>n;
for(int i=0;i<1<<n;i++)
{
for(int j=n-1;j>=0;j--) //从高位枚举到低位
{
if(i>>j&1) //i的二进制的第j位存在,说明选了第j个数
{
cout<<"Y";
}
else //说明第j个没有选
{
cout<<"N";
}
}
cout<<endl;
}
return 0;
}