GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1567    Accepted Submission(s): 751

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

题目大意：

数据量太大，用常规方法做是行不通的。后来看了别人的解题报告说，先找出N的约数x，

          并且gcd（x，N）>= M，结果为所有N/x的欧拉函数之和。

          因为x是Ｎ的约数，所以ｇｃｄ（ｘ，Ｎ）＝ｘ >= M；

　　　设ｙ＝Ｎ／ｘ，ｙ的欧拉函数为小于ｙ且与ｙ互质的数的个数。

　　　设与ｙ互质的的数为ｐ１，ｐ２，ｐ３，…，ｐ４

　　　那么ｇｃｄ（ｘ* pi，N）= x >= M。

          也就是说只要找出所有符合要求的y的欧拉函数之和就是答案了。

至于为何用ans+=Euler(n/i0而不是直接加n/i，这便是为了查重，防止出现重复

只加满足gcd(k*x,n)==x的个数，不过如果直接遍历找约数遍历的量太大，同样会

TLE，因此可以把区间右坐标变成sqrt(n)，同时判断i和n/i即可，特别注意num*num=n

的情况要单独处理。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
int kase=0;
LL Euler(LL n)
{
    LL ans=n;
    for(int 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()
{
    int t;
    cin>>t;
    __int64 n,m;
    while(t--)
    {
        kase=0;
        scanf("%I64d%I64d",&n,&m);
        int num=(int)sqrt(n+0.5);
        //cout<<num<<endl;
        for(int i=1;i<num;i++)
        {
            if(n%i==0)
            {
                if(n/i>=m)
                kase+=Euler(i);
                if(i>=m)
                kase+=Euler(n/i);
            }
        }
        //cout<<kase<<endl;
       // cout<<Euler(100)<<endl;
        if(num*num==n&&num>=m) kase+=Euler(num);
        cout<<kase<<endl;
    }
    return 0;
}