hdoj2588GCD【欧拉函数】

最新推荐文章于 2018-08-06 19:47:37 发布

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最新推荐文章于 2018-08-06 19:47:37 发布

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分类专栏：欧拉函数文章标签： hdoj2588GCD欧拉函数

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本文探讨了一种基于欧拉函数的算法，用于解决特定条件下两个正整数最大公约数的问题。该算法通过枚举和计算欧拉函数值来找出与给定整数N的最大公约数大于指定值M的所有整数X。

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GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1331 Accepted Submission(s): 604

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：求1~N中与N的最大公约数大于M的个数

设N=a*b，X=a*d；gcd（N,X）==a所以b和d互质因为X<=N所以d为小于等于b的任意数；所以d的数量即为b的欧拉函数值所以只需要枚举a求出b的欧拉函数值即可

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int Eular(int n){
	int ret=1,i;
	for(i=2;i*i<=n;i++){
		if(n%i==0){
			n/=i;ret*=i-1;
			while(n%i==0){
				n/=i;ret*=i;
			}
		}
	}
	if(n>1)
	ret*=n-1;
	return ret;
}
int main()
{
	int t,i,j,ans,N,M;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&N,&M);
		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;
}