hdu4497 GCD and LCM

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1460    Accepted Submission(s): 653

Problem Description

Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L?
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z.
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions.

 

Input

First line comes an integer T (T <= 12), telling the number of test cases.
The next T lines, each contains two positive 32-bit signed integers, G and L.
It’s guaranteed that each answer will fit in a 32-bit signed integer.

 

Output

For each test case, print one line with the number of solutions satisfying the conditions above.

 

Sample Input

2 6 72 7 33 

 

Sample Output

72 0

 
  给出gcd和lcm，求出满足条件的(x,y,z)这样的三元组有多少个。
  把这些数分解因子，gcd因子个数是这些数里面拥有这个因子最少的数的因子个数，gcd因子个数是这些数里面拥有这个因子最多的数的因子个数。所以这里面先用lcm除以gcd，把除了之后得到的数分解因子，对于每个因子，假设个数是m，x,y,z里面一定有一个数的这个因子个数是m，一个数因子个数为m，剩下一个在0到n之间。因此对于这个因子，三个数的组合有(m-1)*6+6种。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;const int MAXN=300010;const int SIZE=4096;const int INF=0x3f3f3f3f;int T;LL G,L;int main(){    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--){        scanf("%lld%lld",&G,&L);        if(L%G!=0||L<G){            printf("0\n");            continue;        }        LL n=L/G;        LL sq=sqrt(n)+1;        int ans=1;        for(LL i=2;i<=n&&i<=sq;i++){            int m=0;            while(n%i==0){                n/=i;                m++;            }            if(m>0) ans*=(m-1)*6+6;        }        if(n>1) ans*=6;        printf("%d\n",ans);    }    return 0;}