南邮 OJ 1126 GCD

GCD

时间限制(普通/Java) : 2000 MS/ 6000 MS          运行内存限制 : 65536 KByte
总提交 : 141            测试通过 : 23 

比赛描述

The greatest common divisor GCD(a,b) of two positive integer a and b,sometimes written (a,b),
is the largest divisior 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 difficlut problem.
Given integer N and M,how many integer X satisfies 1<=X<=N and (X,N)≥M.

输入

The first line or input is an integer T(<=3000)representing the number of test cases.The following T lines each contains two numbers N and M(1<=N<=1000000000,0<=M<=1000000000),representing a test case.

输出

For each test case,output the answer on a single line.

样例输入

3
1 1
10 2
10000 72

样例输出

1
6
260

题目来源

第九届中山大学程序设计竞赛预选题

/*    题意：求有多少x(1<=x<=n),使得gcd(x,n)>=m;    先求n的所有大于等于m的因子,ei    答案ans=∑phi[n/ei];phi[i]为欧拉函数，为不大于i且与i互质的正整数个数    why?    对于一个与ei互质且小于等于n/ei的正整数p来说，p*ei<=n,gcd(p*ei,n)=ei;则phi[n/ei]就是1~n中的与n最大公约数是ei的个数。而n与1~n的最大公约数必定是n的因子。所以符合gcd(x,n)>=m的x为n所有大于等于m因子的倍数，用phi即可避免重复。*/#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <iostream>#include <algorithm>using namespace std;const int maxn=1000;int e[maxn];int euler_phi(int n){    int m=(int)sqrt(n+0.5);    int ans=n,i;    for(i=2;i<=m;i++)    {        if(n%i==0)        {            ans=ans/i*(i-1);            while(n%i==0)n/=i;        }    }    if(n>1)ans=ans/n*(n-1);    return ans;}int main(){    int T,n,m;    cin>>T;    while(T--)    {        cin>>n>>m;        int i,j,k,t=0,num,ans=0;        num=(int)sqrt(n+0.5);        for(i=1;i<num;i++)        {            if(n%i==0)            {                if(i>=m)e[t++]=n/i;                if(n/i>=m)e[t++]=i;            }        }        if(num*num==n&&num>=m)e[t++]=num;        for(i=0;i<t;i++)            ans+=euler_phi(e[i]);        cout<<ans<<endl;    }    return 0;}