hdoj-【2588 GCD】

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1895    Accepted Submission(s): 949

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
#include<cstdio>typedef long long LL;LL euler(LL n){LL i,ans=n;for(i=2;i*i<=n;++i){if(n%i==0){ans=ans*(i-1)/i;while(n%i==0)n/=i; } }if(n!=1)ans=ans*(n-1)/n;//printf("%lld ",ans); return ans;} int main(){int t;scanf("%d",&t);while(t--){LL n,m,sum=0,i;scanf("%lld%lld",&n,&m);for(i=1;i*i<=n;++i){if(n%i)continue;if(i>=m)sum+=euler(n/i);if(n/i>=m&&i*i!=n)sum+=euler(i); } printf("%lld\n",sum); } return 0;}