hdu2588 GCD 给定n，m。求x属于[1,n]。有多少个x满足gcd(x,n)>=m; 容斥或者欧拉函数

GCD
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2155    Accepted Submission(s): 1093


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



欧拉函数做法：
/**
题目：GCD
链接：http://acm.hdu.edu.cn/showproblem.php?pid=2588
题意：给定n，m。求x属于[1,n]。有多少个x满足gcd(x,n)>=m;
思路：
x -> [1,n]

d = gcd(x,n) >= m

d肯定为n的约数。

对一个确定的d = gcd(x,n);

那么：gcd(x/d,n/d) = 1;

满足上面式子的x为：f(n/d); f(y)表示y的欧拉函数。

sigma(f(n/d)) (d为n的约数且d>=m);

f(y) = y*(p1-1)/p1*(p2-1)/p2...*(pe-1)/pe;

*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#define LL long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
ll Euler(ll x)
{
    ll n = x;
    for(int i = 2; i*i<=x; i++){
        if(x%i==0){
            n = n/i*(i-1);
            while(x%i==0)x/=i;
        }
    }
    if(x>1){
       n = n/x*(x-1);
    }
    return n;
}
vector<int> v;
int main()
{
    int T;
    int n, m;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        v.clear();
        for(int i = 1; i*i<=n; i++){
            if(n%i==0){
                if(i*i==n){
                    if(i>=m) v.push_back(i);
                }
                else{
                    if(i>=m) v.push_back(i);
                    if(n/i>=m) v.push_back(n/i);
                }
            }
        }
        int len = v.size();
        int cnt = 0;
        for(int i = 0; i < len; i++){
            cnt += Euler(n/v[i]);
        }
        printf("%d
",cnt);
    }
    return 0;
}

容斥做法：
/**
题目：GCD
链接：http://acm.hdu.edu.cn/showproblem.php?pid=2588
题意：给定n，m。求x属于[1,n]。有多少个x满足gcd(x,n)>=m;
思路：

显然d = gcd(x,n)中的d一定是n的约数。

显然d = gcd(d,n); 先获得所有>=m的d;

那么d的倍数为x=k*d，如果小于等于n，则一定也满足gcd(x,n)>=m；

k = n/d; 如果对每个d这样计算，会有重复的计算。

当d = 2, 3时候，x=6会多计算一次。

所以要对所有的d进行容斥处理。

问题转化为：n的约数为d，求解d>=m的所有的d在n范围内至少有一个是d的倍数的数有多少个。



*/
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#define LL long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
vector<int> v;
int a[10004], z;  ///注意a数组要开大些，约数个数还是不是100就够的。
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
ll rc(ll n)
{
    ll sum = 0;
    ll mult;
    int ones;
    int len = v.size();
    int m = (1<<len);
    //奇加偶减
    for(int i = 1; i < m; i++){
        ones = 0;
        mult = 1;
        for(int j = 0; j<len; j++){
            if(i&(1<<j)){
                ones++;
                mult = mult/gcd(mult,v[j])*v[j];
                if(mult>n) break;
            }
        }
        if(ones%2==0){
            sum -= n/mult;
        }else
        {
            sum += n/mult;
        }
    }
    return sum;
}
int main()
{
    int T;
    int n, m;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        v.clear();
        z = 0;
        for(int i = 1; i*i<=n; i++){
            if(n%i==0){
                if(i*i==n){
                    if(i>=m) a[z++] = i;
                }
                else{
                    if(i>=m) a[z++] = i;
                    if(n/i>=m) a[z++] = n/i;
                }
            }
        }
        ///出去包含的，比如2,4那么4要去掉。以为4的倍数一定是2的倍数。
        sort(a,a+z);
        for(int i = 0; i < z; i++){
            int sign = 0;
            for(int j = 0; j < i; j++){
                if(a[i]%a[j]==0){
                    sign = 1; break;
                }
            }
            if(sign==0){
                v.push_back(a[i]);
            }
        }
        printf("%lld
",rc(n));
    }
    return 0;
}