C - GCD【化简+欧拉定理】

Discription
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
3
1 1
10 2
10000 72

Sample Output
1
6
260

题意
给定一个N和一个M。
求解在1<=X<=N 的条件下，
使(X,N)>=M成立的X的个数

思路
设X=k*a，N=k*b，k为X和N的公因数；
X成立的个数就是a成立的个数，a<=b；
当k一定时，a成立的个数为小于a且与a互质的数的个数。
枚举k求出在不同的情况下成立的a的个数，累加求和即可

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
///直接求解一个数n的欧拉函数
int euler(int n)  //返回euler(n)
{
    int res=n,a=n;
    for(int i=2; i*i<=a; i++)
    {
        if(a%i==0)
        {
            res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出
            while(a%i==0)
                a/=i;
        }
    }
    if(a>1)
        res=res/a*(a-1);
    return res;
}
int t;
int n,m,ans;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        ans=0;
        for(int i=1;i<=sqrt(n);i++)
        {
            if(n%i==0)
            {
                if(i>=m)
                    ans+=euler(n/i);
                if(n/i>=m&&i*i!=n)
                    ans+=euler(i);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

来源：CSDN

作者：爱吃老谈酸菜的DV

链接：https://blog.csdn.net/weixin_43460224/article/details/103995744