数学训练—-数论HDU 2588 – GCD

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

分析:
设X与N的最大公约数是k
则:x=k*p;N=k*q;(p,q为整数)

进一步分析:p,q是互质的,因为k是其两者的最大的公约数。同时p,q均为质数,

题目求:k>=m的个数,

因为k=N/q>=m 得:q<=N/m,q=N/k;

即题转化为求q得个数,(因为每一个k对应一个q)

因为q为质数

即题可转化为小于N/K的质数的个数

  其中:k>=m

於是调用欧拉函数,求得所得结果

代码:

#include <iostream>

using namespace std;
int oula(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 main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(n%i) continue;
            if(i>=m)
            ans+=oula(n/i);
        }
        cout<<ans<<endl;

    }
    return 0;
}

但是,代码超时了。可以在循环里减小时间复杂度。
代码:

for(int i=1;i*i<=n;i++)
{
      if(n%i) continue;
      if(i>=m&&i*i!=n)
      ans+=oula(n/i);
      if(n/i>=m)
      ans+=oula(i);
 }

改进后的代码就可以过了!!!
加油!

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