HDU

GT and numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1846    Accepted Submission(s): 502

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≤263.

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

 

Source

BestCoder Round #60

 

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 题意：

给你两个数N和M，求是否能进行操作使N乘上他的因子与M相等。如果可以输出最小的操作次数，否则输出“-1”。

思路：

如果N可以得到M，说明M的因子都包等于N的因子中。M中去掉N剩下的因子和N的最大公因式为min（N， M/N）。这样每次乘以最大公因式，可以保证每次乘的因子是最大的。

 

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef unsigned long long LL; // long long会爆
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const long long MOD = 1e9 + 7;
const int MAXN = 100 + 10;
LL gcd(LL x, LL y) {return y? gcd(y, x%y): x;}
int main() {
    int T;
    LL N, M, t;
    scanf("%d", &T);
    while (T--) {
        scanf("%I64u%I64u", &N, &M);
        int ans = 0; t = M;
        while (N != M) {
            LL d = gcd(N, M/N);
            if (N > M || d == 1) {ans = -1; break;}
            N *= d; t /= d;
            ans++;
        }
        printf("%d
", ans);
    }
    return 0;
}