hdoj2588 GCD欧拉函数

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1321    Accepted Submission(s): 598

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

 

Source

ECJTU 2009 Spring Contest

 

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题意：是让求x从1到n,gcb(x,n)>=m个数

由于gcb(a*b,a*c)=a,则b和c互质，假设n>=c*b,gcb(c*b,n)=c,c>=m,这道题就变成求b的欧拉个数.由于数据量太大需要优化。

#include<stdio.h>//求小于x与x互质的个数 int euler(int x){int s=x;for(int i=2;i*i<=x;i++){if(x%i==0){s=s/i*(i-1);while(x%i==0)x/=i;}}if(x>1)s=s/x*(x-1);return s;}int main(){int t,i,ans,n,m;scanf("%d",&t);while(t--){ans=0;scanf("%d%d",&n,&m);for(i=1;i*i<=n;i++)//优化代码 {if(n%i==0){if(i>=m){ans+=euler(n/i);}if(n/i>=m&&i*i!=n)ans+=euler(i);}}printf("%d\n",ans);}return 0;}