題目

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
3
1 1
10 2
10000 72

Sample Output
1
6
260

題目的大致意思是：給定n和m，求在所有小於n的正整數中，有幾個和n的gcd大於等於m。多組資料

題解

     - 如果gcd(a,b)=c   那麼gcd(a/c,b/c)=1
     - 此時求的是對於一個數x，首先它要大於m，那麼只需要gcd(n,x*q)=x，q是正整數且x*q<=n
     - gcd(n,x*q)=x等價於gcd(n/x,q)=1
     - 因此只需要用尤拉函式算出與n/x互質的數有多少即可，同時可以避免重複

程式碼

#include<iostream>
#include<cstdio>
 

#include<cstring>
#include<algorithm>
using namespace std;

int T;
int n,m;
int ans;

int euler(int x){//計算某個數的尤拉函式值
    int tmp=1,i;
    for(i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            x/=i;tmp*=i-1;
            while(x%i==0)
            {x/=i;tmp*=i;}
        }
    }
    if(x>1)
    tmp*=x-1;
    return tmp;
}

int main(){
    scanf("%d",&T);
    while(T--){
        ans=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i*i<=n;++i){
            if(n%i==0){
                if(i>=m) ans+=euler(n/i);
                if(i*i!=n&&n/i>=m) ans+=euler(i);//兩步一起寫可以少迴圈。i*i!=n 用於判重
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}