HDOJ 2588 GCD

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2588

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 218    Accepted Submission(s): 66

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
   
   
   
   

 

 

Source

ECJTU 2009 Spring Contest

 

 

Recommend

lcy

 

这个题数据量比较大，所以暴力是肯定要超时的。故，我们需要思考一些算法来优化我们的程序

我们可以在sqrt(n)的时间内找到n的所有约数，而这些约数和n的最大公约数就是这个约数本身

 

我们可以寻找x>=m且x是n的约数，最后答案就是每一个大于m的约数x

ans=sigma(eular(n/x))

 

可以大概的说一下为什么这样处理

因为对于x的倍数和n取最大公约数，则其值完全可能大于x。

那么可以证明出，必然在m<=k*x<=n之间一定存在eular(n/x)个数与n的最大公约数是x

 

我的代码：
#include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> using namespace std; int a[40000]; int eular(int n) { int ret=1,i; for(i=2;i*i<=n;i++) { if(n%i==0) { n=n/i; ret=ret*(i-1); while(n%i==0) { n=n/i; ret=ret*i; } } if(n==1) break; } if(n>1) ret=ret*(n-1); return ret; } int main() { int i,num,n,m,T,ans; scanf("%d",&T); while(T--) { num=0,ans=0; scanf("%d%d",&n,&m); for(i=1;i*i<n;i++) { if(n%i==0) { a[num++]=i; a[num++]=n/i; } } if(i*i==n) a[num++]=i; sort(a,a+num); for(i=0;i<num;i++) { if(a[i]>=m) ans=ans+eular(n/a[i]); } printf("%d/n",ans); } return 0; }