【数论】（欧拉函数）GCD HDU2588
描述
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.
输入
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.
输出
For each test case,output the answer on a single line.
样例输入
3
1 1
10 2
10000 72
样例输出
1
6
260
分析思路：首先，我们假设X是满足gcd(X，N)=a并且a>=m,则gcd(X/a,N/a)=1。也就是说，找到多少个X/a与N/a互质（典型的欧拉函数应用），就找到多少个X满足题目要求。因为a是不确定的，但是可以确定a是N的因子，所以我们枚举所有因子，然后加上这些因子的欧拉函数即可。
#include <iostream> #include <cstdio> using namespace std; int phi(int n){ int rea=n; for(int i=2;i<=n;i++){ if(n%i==0){ rea=rea-rea/i; do n/=i; while(n%i==0); } } return rea; } int main(){ int t; cin>>t; while(t--){ int n,m; cin>>n>>m; if(m==1){ cout<<"1"<<endl; } else{ int ans=0; for(int i=2;i*i<=n;i++){ if(n%i==0){ if(i>=m&&i*i!=n) ans+=phi(n/i); if(n/i>=m) ans+=phi(i); } } cout<<ans+1<<endl; } } }View Code 优质内容筛选与推荐>>
1、Java线程—中断技术
2、LeetCode 700 Search in a Binary Search Tree 解题报告
3、实验三实验报告
4、经典SQL语句大全
5、平凡的人，不一样的幸福