UVa11889

11889 Benefit
Recently Yaghoub is playing a new trick to sell some more. When somebody gives him A Tomans, he
who never has appropriate changes, asks for B Tomans such that lowest common multiple of A and B
equals to C and he will pay back a round bill. Or otherwise take some snack instead of the remaining of
his money. He believes that finding such a number is hard enough that dissuades students from paying
that.
You should write a program that help poor students giving the appropriate amount of money to
Yaghoub. Of course if there are several answers you go for students’ benefit which is the lowest of them.
Input
The first line begin with an integer T (T  100000), the number of tests. Each test that comes in a
separate line contains two integers A and C (1  A;C  107).
Output
Print the lowest integer B such that LCM(A;B) = C in a single line. If no such integer exists, print
‘NO SOLUTION’ instead. (Quotes for clarity)
Sample Input
32
6
32 1760
7 16
Sample Output
3
55
NO SOLUTION

题意：

       输入两个整数A、C，求最小的B使得lcm(A,B)=C。如果无解则输出“NO SOLUTION”。

分析：

       考虑B=C/A。设G=gcd(A,B)。如果G=1则答案就是B，否则的话考虑一下A和B的素因子唯一分解，A和B的G值一定是它们共同具有的素因子的若干次幂的乘积，我们只需要将数A中具有的这些素因子的幂次全部为B乘上，得到一个新数B’，就是答案。

#include <cstdio>
int gcd(int x,int y){return y == 0 ? x : gcd(y,x % y);}
int main(){
    int T; scanf("%d",&T);
    while(T--){
        int a,c; scanf("%d%d",&a,&c);
        if(c % a) printf("NO SOLUTION
");
        else{
            int b = c / a;
            int g = gcd(a,b);
            if(g == 1){
                printf("%d
",b);
            }
            else{
                int k = 1;
                while(g != 1){
                    k *= g;
                    a /= g;
                    g = gcd(a,b);
                }
                printf("%d
",b * k);
            }
        }
    }
    return 0;
}