[原创]HDU 2588 GCD [欧拉函数]【数论】

2016-08-11 20:25:10 Tabris_ 阅读数:457


博客爬取于2020-06-14 22:43:51
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52186214


题目连接 : 传送阵
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GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1734 Accepted Submission(s): 852

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 :不用解释了吧 …

解题思路:
这道题所需要的算法主要为欧拉函数的运用和一点点的GCD知识。
问题所要求的是 gcd( x , n ) > m ,由gcd( x , n )本身可知,gcd求出来的是 x 和n的最大公约数(设为a),即有式子gcd( x ,n )=a , 进一步进行化简可变为gcd( x/a , n/a )=1 , 到了此处这个式子又有了另一层含义——x/a与n/a互素 。在联想到欧拉函数的功能——对正整数n,欧拉函数是小于或等于n的数中与n互质的数的数目。于是将欧拉函数里的n换成n/a,不就正好能求出x/a的个数了吗?x/a的个数不就是我们所要求的x的个数了吗?

转自这里

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;
}