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 = 1000000;
int prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= MAXN;i++)
    {
        if(!prime[i])prime[++prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= MAXN/i;j++)
        {
            prime[prime[j]*i] = 1;
            if(i % prime[j] == 0)break;
        }
    }
}
long long factor[100][2];
int fatCnt;
int getFactors(long long x)
{
    fatCnt = 0;
    long long tmp = x;
    for(int i = 1; prime[i] <= tmp/prime[i];i++)
    {
        factor[fatCnt][1] = 0;
        if(tmp % prime[i] == 0 )
        {
            factor[fatCnt][0] = prime[i];
            while(tmp % prime[i] == 0)
            {
                factor[fatCnt][1] ++;
                tmp /= prime[i];
            }
            fatCnt++;
        }
    }
    if(tmp != 1)
    {
        factor[fatCnt][0] = tmp;
        factor[fatCnt++][1] = 1;
    }
    return fatCnt;
}
int a[100],b[100];
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 = 0;i < cnt1;i++)
        {
            mp1[factor[i][0]] = factor[i][1];
            a[i] = factor[i][0];
            //printf("%I64d %I64d
",factor[i][0],factor[i][1]);
        }
        getFactors(n);
        int cnt2 = fatCnt;
        bool flag = true;
        for(int i = 0;i < cnt2;i++)
        {
            mp2[factor[i][0]] = factor[i][1];
            if(mp1[factor[i][0]] < factor[i][1])
                flag = false;
            b[i] = factor[i][0];
            //printf("%I64d %I64d
",factor[i][0],factor[i][1]);
        }
        if(!flag)
        {
            printf("0
");
            continue;
        }
        int ans = 1;
        for(int i = 0;i < cnt1;i++)
        {
            int num1 = mp1[a[i]];
            int num2 = mp2[a[i]];
            if(num1 == num2)
                ans *= 1;
            else
            {
                long long tmp = (num1-num2+1)*(num1-num2+1)*(num1-num2+1);
                tmp -= 2*(num1-num2)*(num1-num2)*(num1-num2);
                tmp += (num1-num2-1)*(num1-num2-1)*(num1-num2-1);
                ans *= tmp;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}