View Code of Problem 31
#include<iostream>
using namespace std;
//3 5 7 ==>105(最小公倍数)
//23+105*n
int main() {
long long k;
while(cin>>k) {
cout<<23+(k-1)*105<<endl;
}
} |
Double click to view unformatted code.
Back to problem 31
#include<iostream>
using namespace std;
//3 5 7 ==>105(最小公倍数)
//23+105*n
int main() {
long long k;
while(cin>>k) {
cout<<23+(k-1)*105<<endl;
}
} |
Double click to view unformatted code.