hdu2588——GCD(欧拉函数)
hdu2588——GCD
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
思路:
题意大概:给定N,M(2<=N<=1000000000, 1<=M<=N), 求1<=X<=N 且gcd(X,N)>=M的个数。
解法:数据量太大,用常规方法做是行不通的。后来看了别人的解题报告说,先找出N的约数x,并且gcd(x,N)>= M,结果为所有N/x的欧拉函数之和。因为x是N的约数,所以gcd(x,N)=x >= M;
设y=N/x,y的欧拉函数为小于y且与y互质的数的个数。设与y互质的的数为p1,p2,p3,…,p4。那么gcd(x* pi,N)= x >= M。也就是说只要找出所有符合要求的y的欧拉函数之和就是答案了。
关于欧拉函数求值的知识点总结:(https://www.cnblogs.com/PJQOOO/p/3875545.html);
#include<iostream>
using namespace std;
int euler(int n) {
int res=n,a=n;
for(int i=2; i*i<=a; i++) {
if(a%i==0) {//第一次找到的i必为素因子
res=res/i*(i-1);//res=res-res/i
while(a%i==0) a/=i;//把该素因子全部约掉,相当于i的次方全部约掉
}
}
if(a>1) res=res/a*(a-1);//res=res-res/a;
return res;
}
int main() {
int n,m,t;
scanf("%d",&t);
while(t--) {
scanf("%d %d",&n,&m);
int sum=0,x;
for(int i=1; i*i<=n; i++) {
if(n%i==0) {
if(i>=m) sum+=euler(n/i);
if(n/i>=m&&i!=n/i) sum+=euler(n/(n/i));////注意平方根不能重复算
}
}
printf("%d\n",sum);
}
return 0;
}