hdu 2588 GCD【欧拉+gcd推导*经典】

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2312    Accepted Submission(s): 1167

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

思路：（大神题解）

直接计算绝对超时，所以要想到采用一些定理来进行优化。

①我们先看两个数  N = a*b，X= a*d。因为gcd ( N , X ) = a  所以b,d这两个数互质。又因为d可以是任何一个小于b的数。所以d值数量的的多少就是b的欧拉函数值。所以，我们可以枚举a，然后去求b，然后再求b的欧拉函数值。

②但是如果单纯这样全部枚举的话依旧会超时，所以我们要想一个办法去优化它。我们可以折半枚举，这里的折半并不是二分的意思。

我们先看，我们枚举时，当i<sqrt(n),假设a=n / i， 当i>sqrt(n)之后 有b=n/i，我们观察到当n%i==0时，会出现一种情况，就是a*b==n。所以我们就可以只需要枚举sqrt(n)种情况，然后和它对应的情况就是 n/i。

我们这种枚举时间会快非常多

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int t,n,m;int euler(int a){int res=a;for(int i=2;i*i<=a;i++){if(a%i==0){res=res/i*(i-1);while(a%i==0)a/=i;}}if(a>1) res=res/a*(a-1);return res;}int main(){scanf("%d",&t);while(t--){int ans=0;scanf("%d %d",&n,&m);for(int i=1;i<=sqrt(n);i++){if(n%i==0){if(i>=m) ans+=euler(n/i);//左端if(n/i>=m && i*i!=n) ans+=euler(i);//右端}}printf("%d\n",ans);}return 0;}