NYOJ 题目998 Sum（欧拉函数，水）

最新推荐文章于 2017-07-30 20:19:27 发布

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本文探讨了在给定整数N和M的情况下，如何计算满足条件1<X<=N且(X,N)>=M的整数X之和，并通过提供算法实现来解决此问题。

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GCD

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 3

描述

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,please answer sum of X satisfies 1<=X<=N and (X,N)>=M.

输入

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 (1<=N<=10^9, 1<=M<=10^9), representing a test case.

输出

Output the answer mod 1000000007

样例输入

样例输出

1
35
1305000

上传者

ACM_张书军

丫的今天做题不顺，深夜水道水题压压惊

ac代码

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