#include<bits/stdc++.h> using namespace std; long long n,s; int main(){

cin >> n;
s=n*n;
int k=n,i=1;
while(k!=0){
	k = k/10;
	i = i*10;
}
if((s-n)%i==0){
	cout << "YES";
}
else{
	cout << "NO";
}

return 0; }

双倍经验

#include<bits/stdc++.h>
#define int unsigned long long
using namespace std;
signed main(){
	int n;
	cin>>n;
	int ans1=log10(n)+1,ans2=n*n;
	while(ans1--){
		if(ans2%10!=n%10){
			cout<<"NO";
			exit(0);
		}
		n/=10;
		ans2/=10;
	}
	cout<<"YES";
}

好好好 这不就是前面的自守数吗：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	long long n;
	cin>>n;
	long long qwq=n;
	long long num=1;
	while(qwq>0)
	{
		qwq/=10;
		num*=10;
	}
	if(n*n%num==n)
	{
		cout<<"YES";
	}
	else
	{
		cout<<"NO";
	}
	return 0;
}

C++ :

#include<bits/stdc++.h>
using namespace std;
long long  n,m,l;
int a;
int main()
{
cin>>n;
for(m=1;m<=n;m*=10);
l=n*n;
l=l%m;
if(l==n)
	cout<<"YES";
else
	cout<<"NO";	
return 0;
}

Python :

# coding=utf-8
n=int(input())
if 1<n<10:
    t=n*n
    if t%10==n:
        print("YES")
    else:
        print("NO")

if 10<n<100:
    t=n*n
    if t%100==n:
        print("YES")
    else:
        print("NO")
if 100<n<1000:
    t=n*n
    if t%1000==n:
        print("YES")
    else:
        print("NO")
if 1000<n<10000:
    t=n*n
    if t%10000==n:
        print("YES")
    else:
        print("NO")
if 10000<n<100000:
    t=n*n
    if t%100000==n:
        print("YES")
    else:
        print("NO")
if 100000<n<1000000:
    t=n*n
    if t%1000000==n:
        print("YES")
    else:
        print("NO")
if 1000000<n<10000000:
    t=n*n
    if t%10000000==n:
        print("YES")
    else:
        print("NO")
if 10000000<n<100000000:
    t=n*n
    if t%100000000==n:
        print("YES")
    else:
        print("NO")

显然重题了。而且题面有误。

#include<bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f
using namespace std;
int n;
signed main(){
	cin>>n;
	int s=log10(n)+1;//n的位数
	int m=n*n;
	while(s--){
		if(m%10!=n%10){
			cout<<"NO";
			return 0;
		}
		m/=10;n/=10;
	}
	cout<<"YES";
	return 0;
}