标签：nta iss ring fit case com bottom desc 乘法
GCD and LCM
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3409 Accepted Submission(s):
1503
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
Sample Output
翻译：给出g和l，求满足gcd(x,y,z)=g,lcm(x,y,z)=l的(x,y,z)组合的数量，(1, 2, 3)和(1, 3, 2)算作不同组合
分析：
g是因数，l是倍数，显然可以得到x%g=y%g=z%g=0=l%x=l%y=l%z;
进一步推出l%g=0;若不符合，无解。
根据唯一分解定理
将x，y，z分解
x=p1^a1*p2^a2……ps^as;
y=p1^b1*p2^b2……ps^bs;
z=p1^c2*p2^c2……ps^cs;
g是三个数的最大公约数，则g满足g=p1^min(a1,b1,c1)……ps^min(as,bs,cs)=p1^e1……ps^es
l是三个数的最小公倍数，则l满足l=p1^max(a1,b1,c1)……ps^max(as,bs,cs)=p1^h1……ps^hs
ei=min(ai,bi,ci) hi=max(ai,bi,bi)
x，y，z三个数分解后，对于每一个质因子，必然有一个指数是ei，一个指数是hi，先将这两个数固定下来
另一个数可以取ei到hi之间的数，包括ei和hi，设为ti
(1)当ti=ei或者ti=hi时，(ei,ei,hi)只有三种组合方式，(ei,hi,hi)也只有三种组合方式
(2)当ti≠ei并且ti≠hi时，(ei,ti,hi)有6种组合方式
(3)当ei=ti=hi时，只有1种组合方式
当ei≠hi时，则每个质因子全部的组合方式有6*(hi-ei)种，组合数用乘法计算结果，当hi=ei，不累乘0
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<vector>
#include<iostream>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
#define ll long long
const int maxx=1e6+5;
int prime[maxx];
int vis[maxx];
ll g,l;
int cnt;
void init()
{
memset(vis,true,sizeof(vis));
vis[0]=vis[1]=false;
cnt=0;
for(int i=2;i<=maxx;i++)
{
if(vis[i])
prime[cnt++]=i;
for(int j=0;j<cnt && prime[j]*i<=maxx;i++)
{
vis[ i*prime[j] ]=false;
if(i%prime[j]==0) break;
}
}
}
int main()///hdu4497
{
init();
int t;
scanf("%d",&t);
while(t--)
{
ll ans=1;
scanf("%lld%lld",&g,&l);
if(l%g)
printf("0\n");
else
{
for(int i=0;i<cnt;i++)
{
int e=0,h=0;
while(g%prime[i]==0)
{
e++;
g=g/prime[i];
}
while(l%prime[i]==0)
{
h++;
l=l/prime[i];
}
if(h-e)
{
ans=ans*(h-e)*6;
}
}
if(g==l)///32位范围内大素数因子只能有一个，l能被g整除证明这个大素数相同，不累乘
;
else if(l>1)///如果g=1，l=大素数，再乘一次6
ans=ans*6;
printf("%lld\n",ans);
}
}
return 0;
}
hdu4497-GCD and LCM-(欧拉筛+唯一分解定理+组合数)
标签：nta iss ring fit case com bottom desc 乘法
原文地址：https://www.cnblogs.com/shoulinniao/p/10357950.html
