HDU 5505 GT and numbers【思维+GCD】

最新推荐文章于 2019-02-08 23:49:00 发布

DTL66

最新推荐文章于 2019-02-08 23:49:00 发布

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分类专栏：思维 3.GCD+LCM

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本文探讨了GTandnumbers问题的解决策略，该问题要求计算通过特定操作使数字N至少等于M所需的步骤数。文章详细介绍了如何使用最大公约数(GCD)进行优化，并分享了解题过程中的挑战及心得。

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GT and numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1839 Accepted Submission(s): 500

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 is the test number.

In the next

T lines there are two numbers

N and

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

Sample Output

0
-1
1

一碰到这题就不会了，先写感觉没有错，最后发现题意没理解对,N是一直变得，因子也会变，写了又超时，又忘了优化一下GCD，GCD竟然也会超时，搜一下简化版的GCD，提交一下又MLE，这真是一道好题，那一点想不到都不行，必须优化到最优的形式，还是多思考吧；

#include<cstdio>

typedef unsigned long long LL;

LL GCD(LL a,LL b)
{
	return !b?a:GCD(b,a%b);
} 
int main()
{
	LL T,M,N,i;
	scanf("%llu",&T);
	while(T--)
	{
		bool falg=true; LL cnt=0;
		scanf("%llu %llu",&N,&M);
	    if(N>M||M%N) {
	    	falg=false; 
		}
		else if(N==M) falg=true;
		else {
		     while(N<M)
		     {
		     	 LL tem=GCD(M/N,N);
		     	 if(tem==1)
		     	 {
		     	    falg=false; break;
				 }
				N*=tem; ++cnt; 
			 }
		}
		if(falg) printf("%llu\n",cnt);
		else printf("-1\n");
	}
	return 0;
}