#include <bits/stdc++.h>
using namespace std;
int main()
{
    int m,k;
    int sum = 0;
    cin >> m >> k;
    int num = m;
    while(num>0)
    {
        if(num%10 == 3)
            sum += 1;
        num /=10;//分离数位
    }
    if(m % 19 == 0&&sum == 3)
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int m,k,l;
int main()
{
	cin>>m>>k;
    for(int i=1;i<=100000;i*=10){
        int x=m/i%10;
        if(x==3) l++;
    }
    if(m%19==0&&l==k) cout<<"YES";
    else cout<<"NO";
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
	int M,K;
	cin >> M >> K;
	int count = 0;
	if(M%19==0)  //如果M%19=0,进行下一步判断 
	{
		while(M>0)
		{
			if (M%10==3)
			{
				count++;
			}
            M/=10;
		}
		if(count==K)  //如果count=K，输出YES 
		{
			cout << "YES";
		}
		else  //否则输出NO 
		{
			cout << "NO";
		}
	}
	else  //如果不是，直接输出NO 
	{
		cout << "NO"; 
	}
}

利用sscanf()双向赋值

用char a记录'k'的个数

用int q记录能否整除19

#include <bits/stdc++.h>
int main(){
    int q,k,sum;char a[10];std::cin>>a>>k;int n=strlen(a);
    for(int i=0;i<n;i++) sum+=(a[i]=='3'?1:0);
    sscanf(a,"%d",&q);
    if(q%19==0&&sum==k) std::cout<<"YES";
    else std::cout<<"NO";
    return 0;}

#include<iostream>
#include<iomanip>
using namespace std;
int main(){
	int n,m;
	cin>>n>>m;
	int t = 0;
	if(n%19 == 0){
		while(n!=0){
			if(n%10 == 3)  t++;
			n/=10;
		}
		if(t == m)  cout<<"YES";
		else cout<<"NO";		
	}
	else
		cout<<"NO";


}