hdu5505-GT and numbers-(贪心+gcd+唯一分解定理)

GT and numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2772    Accepted Submission(s): 688

Problem Description

You are given two numbers N and M.

Every step you can get a new N in the way that multiply N by a factor of N .

Work out how many steps can N be equal to M at least.

If N can't be to M forever,print −1 .

 

 

Input

In the first line there is a number T. T is the test number.

In the next T lines there are two numbers N and M .

T≤1000 , 1≤N≤1000000 ,1≤M≤2^63 .

Be careful to the range of M.

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

 

 

Output

For each test case,output an answer.

 

 

Sample Input

3 1 1 1 2 2 4

 

 

Sample Output

0 -1 1

题意：n要变到m，每次乘一个n的因子数，求最少乘几次

坑1：m需要用无符号long long定义

坑2：n的因子会变，变多变大
用r表示需要乘的总数，择优，每次选r和n的gcd，这样每次乘得多，次数就少，同时更新r和n
如果遇到r！=1，gcd=1证明需要乘的数 包含 n没有的因子，退出。

 

#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 unsigned long long
const int p=10007;
const int maxx=1e6+5;
ll n,m;///坑1：无符号long long才放得下
int step;
ll gcd(ll a,ll b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cin>>n>>m;
        step=0;
        if(n==m) printf("0\n");
        else if(n>m || m%n) printf("-1\n");
        else
        {
            ll r=m/n;///r表示 n还要 乘 一个数 变成m
            ll d;
            bool flag=true;
            while(r!=1)
            {
                d=gcd(r,n);
                if(d==1)///出现这种情况，r!=1,和n没有共同因子，则表示m中含有n中不含有的质因子
                {
                    flag=false;
                    break;
                }
                r=r/d;
                n=n*d;///坑2：每次n会变大，因子会更新
                step++;
            }
            if(flag)
                cout<<step<<endl;
            else printf("-1\n");
        }
    }
    return 0;
}