HDU 4497 GCD and LCM (素数筛选+算术基本定理)

     算术基本定理可表述为：任何一个大于1的自然数 N,如果N不为质数，那么N可以唯一分解成有限个质数的乘积N=P1a1P2a2P3a3......Pnan，这里P1<P2<P3......<Pn均为质数，其中指数ai是正整数。这样的分解称为 N 的标准分解式。最早证明是由欧几里得给出的，现代是由陈述证明。此定理可推广至更一般的交换代数和代数数论。

 

lcm(x,y,z)=k;

gcd(x,y,z)=t;

若：x=a*t;   y=b*t;       z=c*t;

则lcm(a,b,c)=k/t;

若k/t=2^A;

则a,b,c中至少有一个数为2^A,至少有一个数是2^0,另外一个数为2^(0~A);共6*A种情况。

则，若k/t=2^A*3^B*5^C;

a,b,c的情况数为：(6*A)*(6*B)*(6*C);

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1978    Accepted Submission(s): 876

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

 

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

 

#include <iostream>#include <cstring>using namespace std;const int max_num = 1000000;int vis[max_num],prim[max_num];int len;int sum;void is_prim(){len = 0;memset(vis,0,sizeof(vis));memset(prim,0,sizeof(prim));for(int i=2;i<max_num;i++){if(!vis[i]){prim[len++] = i;for(int j=2*i;j<max_num;j+=i){vis[j] = 1;}}}}int getPrim(int nn){int i;int s = 0;sum = 1;for(i=0;i<len && prim[i]*prim[i]<=nn;i++){s=0;while(nn%prim[i]==0){s += 6;nn/=prim[i];}if(s)sum *= s;}if(nn > 1)sum *= 6;return sum;}int main(){int i,j,k,m,n;int t;int g,l;is_prim();cin>>t;while(t--){cin>>g>>l;if(l%g){cout<<"0"<<endl;continue;}cout<<getPrim(l/g)<<endl;}return 0;}