HDU 2588 GCD 【Euler + 暴力技巧】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=2588

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3608    Accepted Submission(s): 1954

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

 

Source

ECJTU 2009 Spring Contest

题意概括：

求 1~N 的范围内存在多少个 X 使得 GCD( X, N ) >= M;

解题思路：

设 s = GCD( X, N);

可知： s >= M, 

且存在 a, b 使得 s*a = X, s*b = N, GCD( a, b ) = 1;

因为 X <= N 所以 a <= b;

综上所述：

N 1e9 的范围缩小一半枚举 s ,求得 b；（因为可以同时求得 i 和 N/i 的方案数）

即求满足 GCD（a, b) = 1 且 a <= b 的 a 的个数。

AC code:

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

const int MAXN = 1e9+10;
LL N, M;

LL Euler(LL n)
{
    LL res = n;
    for(LL i = 2; i*i <= n; i++){
        if(n%i == 0) res = res/i*(i-1);
        while(n%i == 0) n/=i;
    }
    if(n > 1) res = res/n*(n-1);
    return res;
}

int main()
{
    int T_case;
    scanf("%d", &T_case);
    LL ans, b;
    while(T_case--){
        scanf("%lld %lld", &N, &M);
        ans = 0;
        for(LL s = 1; s*s <= N; s++){
            if(N%s) continue;
            if(s >= M) ans+=Euler(N/s);
            if(s*s != N && N/s >= M) ans+=Euler(s);
        }
        printf("%lld
", ans);
    }
    return 0;
}