HDU 2588 GCD

GCD

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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

 

Source

ECJTU 2009 Spring Contest

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。

 

#include
#include
int Euler(int n)
{
    if(n==1)
        return 1;
    int i=2,m=n,root=(int)sqrt(n);
    while(i<=root)
    {
        if(m%i==0)
        {
            n-=n/i;
            while(m%i==0)
                m/=i;
            root=(int)sqrt(m);
        }
        i++;
    }
    if(m!=1)
    {
        n-=n/m;
    }
    return n;
}
int solve(int n,int m)
{
    int nn = sqrt(n),ans=0;
    for(int i=1;i<=nn;i++)
    {
        if(n%i) continue;
        if(i>=m&&i!=nn)
            ans += Euler(n/i);
        if(n/i>=m)
            ans += Euler(i);
    }
    return ans;
}
int main()
{
    int n,t,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf("%d\n",solve(n,m));
    }
    return 0;
}