hdu 2504 又见GCD

又见GCD

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 22598 Accepted Submission(s): 9312

Problem Description

有三个正整数a,b,c(0

#include<cstdio>

int gcd(int a, int b){
if(b == 0){
return a;
}
return gcd(b, a%b);
}

int main(){
int T;
scanf("%d", &T);
while(T--){
int a, b, c;
scanf("%d%d", &a, &c);
b = 2 * c;
while(gcd(a, b) != c){
b += c;   //b是a、c的最大公约数，因为c！=b，所以c=2*b、3*b……分别代入gcd的公式中判断
}
printf("%d\n", b);
}
return 0;
}