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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case,output the answer on a single line.

[align=left]Sample Input[/align]

3
1 1
10 2
10000 72

[align=left]Sample Output[/align]

1
6
260//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;
}