GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1089 Accepted Submission(s): 498
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.
(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
/*
HDOJ 2588 欧拉函数
给定N,M(2<=N<=1000000000, 1<=M<=N), 求1<=X<=N 且gcd(X,N)>=M的个数。
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，
则: x<ｘ* pi<N, 满足1<=X<=N；
那么ｇｃｄ（ｘ* pi，N）=x>= M。pi是与y质数
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int phi(int x)
{
int i,ans=x;
for(i=2;i*i<=x;i++)
{
if(x%i==0)
{
ans-=ans/i;
while(x%i==0)
{
x=x/i;
}
}
}
if(x>1)
ans-=ans/x;
return ans;
}
int solve(int n,int m)
{
int i,sum=0;
for(i=1;i*i<=n;i++)
{
if(n%i!=0)
continue;
if(i>=m&&i*i!=n)//当i=nn时，n/i=nn，要避免求了两次欧拉函数
sum+=phi(n/i);
if(n/i>=m) //对称的
sum+=phi(i);
}
return sum;
}
int main()
{
int n,m,t;
//freopen("test.txt","r",stdin);
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
printf("%d\n",solve(n,m));
}
return 0;
}