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 3 1 1 10 2 10000 72   Sample Output 1 6 260 #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; }