hdoj 2588 GCD【欧拉函数】

最新推荐文章于 2020-08-06 16:17:21 发布

原创最新推荐文章于 2020-08-06 16:17:21 发布 · 850 阅读

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本文探讨了一种在大规模数据范围内高效求解特定整数区间内与给定整数最大公约数(GCD)大于指定值的问题。通过对欧几里得算法的优化及采用二分查找思想，实现了快速求解，避免了在大数据范围内的超时问题。

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GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1320 Accepted Submission(s): 597

Problem Description

The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6.
(a,b) can be easily found by the Euclidean algorithm. Now Carp is considering a little more difficult problem:
Given integers N and M, how many integer X satisfies 1<=X<=N and (X,N)>=M.

Input

The first line of input is an integer T(T<=100) representing the number of test cases. The following T lines each contains two numbers N and M (2<=N<=1000000000, 1<=M<=N), representing a test case.

Output

For each test case,output the answer on a single line.

Sample Input

Sample Output

1
6
260

题意：计算1-N区间里有多少数和N的GCD是大于M的。

解题思路：

这道题一看到 2<=N<=1000000000,应该想到普通的做法肯定要超时，于是要想方法避免超时，“二分思想” 在这得以利用，左右开弓同时查找可以节省大量时间，并且许多合数都不必考虑，代码也从原来的TLE变成0ms.

#include<stdio.h>
int eular(int n)
{
	int rt = 1,i;
	for( i =2; i*i<=n; i++)
	if(n%i==0)
	{
		n/=i;
		rt=rt*(i-1);
		while(n%i==0)
		{
			n/=i;
			rt*=i;
		}
	}
	if(n >1 ) rt =rt*(n-1);
	return rt;
}
int main()
{
	int t,n,m,x,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		int ans = 0;
		for(i = 1; i*i <= n ;i++)
		{
			if(n%i==0)
			{
				if(i>=m) 
					ans+=eular(n/i);
					
				if(n/i>=m&&i*i!=n) 
					ans+=eular(i);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}