UVA - 11889 Benefit【LCM】

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本文介绍了一种解决特殊货币找零问题的方法，通过编写程序帮助学生计算在特定条件下能够获得的最佳找零金额。该问题涉及到数学中的最小公倍数概念，并通过算法找到满足条件的最小数值。

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
3
2 6
32 1760
7 16

Sample Output
3
55
NO SOLUTION

已知A,C 求最小的B满足LCM(A,B)=C;

AC代码：

#include<cstdio>

typedef long long LL;

LL GCD(LL a,LL b) {
	return  !b?a:GCD(b,a%b);
}
LL LCM(LL a,LL b) {
	return a/GCD(a,b)*b;
}

int main() {
	int T; scanf("%d",&T);
	while(T--) {
		int A,C; scanf("%d%d",&A,&C);
	    int B=C/A;
		if(A*B!=C) printf("NO SOLUTION\n");
		else {
		    for(int i=1; ;++i) {
		    	if(LCM(A,B*i)==C) {
		    	    B*=i; break;	
				}
			} 
			printf("%d\n",B);
		}	
	} 
	return 0;
}