[原创]hdu 2588 GCD [欧拉函数]【数论】*

2016-08-11 11:48:04 Tabris_ 阅读数：345

博客爬取于2020-06-14 22:43:55
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52181264

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2588
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GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1700 Accepted Submission(s): 829

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
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题目大意 : 不解释

题解:

首先明确的是，要求的gcd⁡(x,N)∣N\gcd(x,N) |N成立
那么我们只要知道gcd⁡\gcd结果为N的每一个约数时的个数就行了,
当[d∣N]成立,那么就是求gcd⁡(N,x)=d的个数,显然只gcd⁡(N,d)=d,当且仅当gcd⁡(N,d∗z)=d当[d|N]成立,那么就是求\gcd(N,x)=d的个数,\\显然只\gcd(N,d)=d,当且仅当\gcd(N,d*z)=d

gcd(a,b)>=c 可以转化成gcd(a/c,b/c)=1;

那么gcd⁡(N/d,d∗z/d)=1gcd⁡(N/d,z)=1\gcd(N/d,d*z/d)=1 \\ \gcd(N/d,z)=1

那么就是 求小于N/d的与N/d互质的数的个数即可,也就是欧拉函数小于N/d的与N/d互质的数的个数即可,也就是欧拉函数

最后答案就是∑d∣n,d>=mϕ(d)\sum_{d|n,d>=m}\phi(d)

题目所求的欧拉函数也同理

最后O(sqrt(n)^(3/2))可解决

附本题代码
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# include <stdio.h>
# include <vector>
# include <iostream>
# include <stdlib.h>
using namespace std;
# define LL long long int
# define pb push_back

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;
        if(i>=m&&i*i!=n)
            ans += Euler(n/i);
        if(n/i>=m)
            ans += Euler(i);
    }
    return ans;
}

int main()
{
    int _,p=0;
    scanf("%d",&_);
    while(_--)
    {
        int  n,m;
        scanf("%d%d",&n,&m);
        int sum=solve(n,m);
        printf("%d\n",sum);
    }
    return 0;
}