GCD（欧拉函数）

GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 76 Accepted Submission(s): 50

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

 

Source

ECJTU 2009 Spring Contest

 

Recommend

lcy

/*
题意：给出N，M让你求出 X的个数 ，X满足GCD(X,N)>=M;

初步思路：首先N的比M大的因子肯定是，是这个因子的倍数的也是。除此之外就没了，因为其他的数GCD(other,N)=1,如果M等于1的话，直接输出N就行了
    现在的问题就是怎么找因子的倍数，因为会有重复的，res=N/x(x是N的因子);对于因子x有res个可满足的结果，但是在计算过程中会有重复的存在，
    这样，令pi<=res && GCD(pi,res)==1,这样保证了 pi*x不会重复，就转化成了，求N/x的欧拉函数

*/
#include<bits/stdc++.h>
using namespace std;
/**************************欧拉函数模板*****************************/
//直接求解欧拉函数
int euler(int n){ //返回euler(n)
    int res=n,a=n;
    for(int i=;i*i<=a;i++){
        if(a%i==){
            res=res/i*(i-);//先进行除法是为了防止中间数据的溢出
            while(a%i==) a/=i;
        }
    }
    if(a>) res=res/a*(a-);
    return res;
}
/**************************欧拉函数模板*****************************/
int solve(int n,int m){
    if(m==) return n;
    int cur=;
    for(int i=;i*i<=n;i++){
        if(n%i==){//i是n的因子
            if(i>=m){
                cur+=euler(n/i);
            }
            if(i*i!=n){//对面的因子
                if(n/i>=m){
                    cur+=euler(n/(n/i));
                }
            }
        }
    }
    return cur+;
}
int t;
int n,m;

int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        printf("%d\n",solve(n,m));
    }
    return ;
}