hdoj GCD 2588 （欧拉函数）

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1373    Accepted Submission(s): 633

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

31 110 210000 72

Sample Output

16260
//题意：
让求gcd(x,n)>=m;的数的个数。
#include<stdio.h>#include<string.h>#include<math.h>long long ehpi(long long n){long long ans=n;long long m=sqrt(n+0.5);for(int i=2;i<=m;i++)if(n%i==0){ans=ans/i*(i-1);while(n%i==0)n/=i;}if(n>1)ans=ans/n*(n-1);return ans;}int main(){int t;long long n,m;scanf("%d",&t);while(t--){scanf("%lld%lld",&n,&m);long long cnt=0;for(int i=1;i<=sqrt(n);i++){if(n%i)continue;if(i>=m)cnt+=ehpi(n/i);if(n/i>=m&&i*i!=n)cnt+=ehpi(i);}printf("%lld\n",cnt);}return 0;}