hdu 2588 欧拉函数的应用

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

题解：

自己之前做过一道类似的。
在51nod 1040 求一个数的最大公约数之和
大致思路是
一个因子的贡献是gcd(n,i)=x
我们对这个式子进行一个简单的处理gcd(n/x,i/x)=1
也就是求phi(n/x)*x之和
枚举因子的时候可以做一个简单的优化，但是注意根号n不要计算两次。
此题的思路：
题目要求的是gcd(x,n)>=m
参考51nod1040的思路，其实我们可知gcd(x,n)计算的出来一定是n的因子，假设gcd(x,n)=a，通过转换可得gcd(x/a,n/a)=1，转化为枚举n的因子，求phi(n/a)之和

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int Euler(int n){    if(n==1) return 1;    int m=n;    for(int i=2;i*i<=m;i++)    {        if(m%i==0)        {            n-=n/i;            while(m%i==0)                m/=i;        }    }    if(m!=1)    {        n-=n/m;    }    return n;}int solve(int n,int m){    int ans=0;    for(int i=1;i*i<=n;i++)    {        if(n%i) continue;        int tmp = n/i;        if(i>=m)        ans+=Euler(tmp);        if(i!=tmp&&tmp>=m)        {            ans+=Euler(i);        }    }    return ans;}int main(){    int T,p=0;    scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%d",&n,&m);        int sum = solve(n,m);        printf("%d\n",sum);    }    return 0;}