11388 – GCD LCM

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    long long G,L,a,b;
    int nCases;
    scanf("%d",&nCases);
    while(nCases-->0)
    {
        a=b=1;
        scanf("%lld %lld",&G,&L);

            int factor;
            factor=1;
            bool check;
            check=true;
            while(check&&a<=L)
            {
                a=G*factor;
                factor++;
                b=(G*L)/a;
                if(b%G==0&&L%b==0)
                    check=false;
            }
            if(!check)
            printf("%lld %lld\n",a,b);
            else printf("-1\n");

    }
    return 0;
}

that’s my solution
another solution was discussed in the board:

long G,L;
scanf("%ld %ld",&G,&L);
if(L%G !=0)
printf(“-1″);
else printf(G L)

 

so Simple

10193 – All You Need Is Love

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <math.h>
using namespace std;
int GCD(int a,int b)
{
    if(b==0)
        return a;
    else return GCD(b,a%b);
}
int main()
{
    char s1[50];
    char s2[50];
    int nCases;
    scanf("%d",&nCases);
    getchar();
    int test;
    test=1;
    while(nCases-->0)
    {
        gets(s1);
        int j,len1,len2;
        len1=strlen(s1);

        gets(s2);
        len2=strlen(s2);
        int n1,n2;
        n1=n2=0;
        for(j=0;j<len1;j++)
            if(s1[j]=='1')
            {
                n1=n1+(int)pow(2,len1-j-1);
            }
        for(j=0;j<len2;j++)
            if(s2[j]=='1')
            {
                n2=n2+(int)pow(2,len2-j-1);
            }
        if(GCD(n1,n2)>1)
            printf("Pair #%d: All you need is love!\n",test++);
        else printf("Pair #%d: Love is not all you need!\n",test++);
    }
    return 0;
}