HDU 2588 GCD

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 672    Accepted Submission(s): 295

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
   
   
   
   

 

解题：  eular函数

 

思路： 
/*

把所有的 n 的 约数找出来，然后 n = p * d,那我们现在把每一个最大公约数为 d 的全部找出来，怎样找呢？肯定是 x = q * d,显然 q 得与 p 互质，这样才能保证 n 跟 x 的最大公约数是 d，然后我们就只要把所有的 d 找出来就行了，然后就只要找 p 的质因子就行了，这不就是欧拉吗？
*/

#include<cstdio>
#include<cmath>

using namespace std;

const int MAXN=50000;

__int64 pri[MAXN];
bool hash[MAXN];
__int64 N,M,s;

void Is_pri()
{
	__int64 i,j;
	pri[0]=0;
	for(i=2;i<MAXN;i++)
	{
		if(!hash[i])
		{
			pri[++pri[0]]=i;
			for(j=i*i;j<MAXN;j+=i)
				hash[j]=true;
		}
	}
}

__int64 eular(__int64 n)
{
	__int64 tmp=n,i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			tmp-=tmp/i;
			while(n%i==0)
				n/=i;
		}
	}
	if(n>1)
		tmp-=tmp/n;
	return tmp;
}

void DFS(__int64 x,int flag)
{
	int i;
	if(x>=M)
	{
		s+=eular(N/x);
//		printf("%I64d %I64d\n",x,s);
	}
	for(i=flag;i<=pri[0]&&pri[i]*x<=N;i++)
	{
		if(N%(pri[i]*x)==0)
			DFS(pri[i]*x,i);
	}
}

int main()
{
	int T;
	int i,j;

	scanf("%d",&T);

	Is_pri();

	while(T--)
	{
		scanf("%d%d",&N,&M);

		s=0;
		DFS(1,1);
	
		printf("%d\n",s);
	}

	return 0;
}

 

 

网上找到的，思路和我一样，但是代码更犀利，用的时间更短的代码。

 

#include<iostream> 
#include<cstdio> 
using namespace std; 
const int Max  = 1000000005; 
int num[40000],cnt2; 
int Eular( int n ) 
{ 
    int ret = 1; 
    for(  int i = 2; i * i <= n; ++i ) 
    { 
        int temp = i; 
        if( n % temp == 0 ) 
        { 
            ret *= ( temp - 1 ); 
            n /= temp; 
            while( n % temp == 0 ) 
                ret *= temp,n /= temp; 
        } 
    } 
    if( n > 1 ) 
        ret *= ( n - 1 ); 
    return ret; 
} 
int main(  ) 
{ 
    int i,Cas; 
    scanf( "%d",&Cas ); 
    while( Cas-- ) 
    { 
        int n,m; 
        scanf( "%d%d",&n,&m ); 
        cnt2 = 0; 
        for(i = 1; i * i <= n; ++i ) 
            if( n % i == 0 ) 
            { 
                num[cnt2++] = i; 
                if( i * i < n ) 
                    num[cnt2++] = n / i; 
            } 
        int s = 0; 
        for(i = 0; i < cnt2; ++i ) 
            if( num[i] >= m ) 
                s += Eular( n / num[i] ); 
        printf( "%d\n",s ); 
    } 
    return 0; 
}