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.
(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;
}