#include<bits/stdc++.h>
using namespace std;
int m,n,s;
int main()
{
    scanf("%d%d%d",&m,&n,&s);
    printf("%d",s/n*m);
    return 0;
}

教教我怎么发成你们那样的题解

#include <bits/stdc++.h>
using namespace std;
int l, w, n, s;
int main()
{
    cin >> l >> n >> s;
    w = s / n;
    cout << l * w << endl;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int m,n,s;
int main() {
	cin>>m>>n>>s;
	cout<<(s/n)*m;
	return 0;
}

#include<bits/stdc++.h> using namespace std; int m,n,s; int main() { cin>>m>>n>>s; cout<<(s/n)*m; return 0; }

#include<bits/stdc++.h>
using namespace std;
int m,n,s;
int main() {
	cin>>m>>n>>s;
	cout<<(s/n)*m;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int m,n,s;
    cin>>m>>n>>s;
    cout<<m*(s/n);
    return 0;
}

··· #include<bits/stdc++.h>

using namespace std;

int main(){

int a,b,c;

cin>>a>>b>>c;

cout<<c/b*a;

return 0;

} ···

#include<bits/stdc++.h>

using namespace std;

int main(){

int a,b,c;

cin>>a>>b>>c;

cout<<c/b*a;

return 0;

}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    c = c / b * a;
    cout<<c;
    return 0;
}

简单A

过辣！

    long long m,n,s,b;
    cin>>m>>n>>s;
    b=s/n;//b是原来的宽
    cout<<m*b;