HDU 4497 GCD and LCM （合数分解）

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 40    Accepted Submission(s): 22

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

 

Recommend

liuyiding

 

对G和L 分别进行合数分解。

很显然，如果G中出现了L中没有的素数，或者指数比L大的话，肯定答案就是0了、

对于素数p,如果L中的指数为num1,G中的指数为num2.

显然必须是num1 >= num2;

3个数p的指数必须在num1~num2之间，而且必须有一个为num1,一个为num2

容斥原理可以求得种数是：

(num1-num2+1)*(num1-num2+1)*(num1-num2+1) -

2*(num1-num2)*(num1-num2)*(num1-num2) +

 (num1-num2-1)*(num1-num2-1)*(num1-num2-1);

然后乘起来就是答案：

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/24 12:48:31
 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1005.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 const int MAXN = ;
 int prime[MAXN+];
 void getPrime()
 {
     memset(prime,,sizeof(prime));
     for(int i = ;i <= MAXN;i++)
     {
         if(!prime[i])prime[++prime[]] = i;
         for(int j = ;j <= prime[] && prime[j] <= MAXN/i;j++)
         {
             prime[prime[j]*i] = ;
             if(i % prime[j] == )break;
         }
     }
 }
 long long factor[][];
 int fatCnt;
 int getFactors(long long x)
 {
     fatCnt = ;
     long long tmp = x;
     for(int i = ; prime[i] <= tmp/prime[i];i++)
     {
         factor[fatCnt][] = ;
         if(tmp % prime[i] ==  )
         {
             factor[fatCnt][] = prime[i];
             while(tmp % prime[i] == )
             {
                 factor[fatCnt][] ++;
                 tmp /= prime[i];
             }
             fatCnt++;
         }
     }
     if(tmp != )
     {
         factor[fatCnt][] = tmp;
         factor[fatCnt++][] = ;
     }
     return fatCnt;
 }
 int a[],b[];
 map<int,int>mp1,mp2;
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     getPrime();
     //for(int i = 1;i <= 20;i++)
         //cout<<prime[i]<<endl;
     int n,m;
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         mp1.clear();
         mp2.clear();
         getFactors(m);
         int cnt1 = fatCnt;
         for(int i = ;i < cnt1;i++)
         {
             mp1[factor[i][]] = factor[i][];
             a[i] = factor[i][];
             //printf("%I64d %I64d\n",factor[i][0],factor[i][1]);
         }
         getFactors(n);
         int cnt2 = fatCnt;
         bool flag = true;
         for(int i = ;i < cnt2;i++)
         {
             mp2[factor[i][]] = factor[i][];
             if(mp1[factor[i][]] < factor[i][])
                 flag = false;
             b[i] = factor[i][];
             //printf("%I64d %I64d\n",factor[i][0],factor[i][1]);
         }
         if(!flag)
         {
             printf("0\n");
             continue;
         }
         int ans = ;
         for(int i = ;i < cnt1;i++)
         {
             int num1 = mp1[a[i]];
             int num2 = mp2[a[i]];
             if(num1 == num2)
                 ans *= ;
             else
             {
                 long long tmp = (num1-num2+)*(num1-num2+)*(num1-num2+);
                 tmp -= *(num1-num2)*(num1-num2)*(num1-num2);
                 tmp += (num1-num2-)*(num1-num2-)*(num1-num2-);
                 ans *= tmp;
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }