as down:

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,b,n,yui,yumin=99999999;
	cin>>n;
    for(int i=1;i*69<=n;i+=1)
    {
    	yui=(n-i*69)%39;
    	if(yui<yumin)
    	{
    		yumin=yui;
    		a=i;
    	}
    	
    }
    b=(n-a*69)/39;
    cout<<a<<" "<<b<<" "<<yumin;
	return 0;
}

注意给o的赋值不要太小，也不要太大，7<=o<=10^7,在这个区间任意整数值都可以，具体为什么原理我也不清楚 #include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,n,m,o=10^7;
cin>>n;
for(int i=1;i69<=n;i+=1)
{
m=(n-i69)%39;
if(m<o)
{
o=m;
a=i;
}

}           
b=(n-a*69)/39;             
cout<<a<<" "<<b<<" "<<o;           
return 0;             

}

#include<bits/stdc++.h> using namespace std; int main() { int x,s,a,b=100,c; cin>>x; for(int i=1;i69<=x;i++) { a=(x-i69)%39; if(a<b) { b=a; s=i; } } c=(x-s*69)/39; cout<<s<<" "<<c<<" "<<b<<endl; return 0; }

#include <bits/stdc++.h>
using namespace std;

int main()

{
	int n,shengyu,num_long=0,num_short=0,ans=99999;
	//ans should be big enough.
	
	cin>>n;
	
	
	for(int i=1;i*69<=n;i++)
	{
		
		shengyu=n-i*69-((n-i*69)/39)*39;
		if(shengyu<ans)
		{
			ans=shengyu;
			num_long=i;
			num_short=(n-i*69)/39;
		}
		
	}
	cout<<num_long<<" "<<num_short<<" "<<ans;
	return 0;
}