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+;
 LL N, M;

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

 int main()
 {
     int T_case;
     scanf("%d", &T_case);
     LL ans, b;
     while(T_case--){
         scanf("%lld %lld", &N, &M);
         ans = ;
         for(LL s = ; 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\n", ans);
     }
     return ;
 }