hdu 2588 GCD (欧拉函数)

GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 890 Accepted Submission(s): 396

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

1 1

10 2

10000 72

Sample Output

260

Source

ECJTU 2009 Spring Contest

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 1 //15MS    208K    784 B    G++
 2 /*
 3  
 4    题意：
 5        求[1,n]内与n公约数大于m的数的个数。
 6        
 7    思路：
 8        想了挺长时间，没什么思路。
 9        后来得知结果为：
10            对于所有大于m的n的公约数k，求出所有n/k的欧拉数的和即为题解。
11        一个数m的欧拉数即为小于m且与m互质的数的和。
12        至于怎么得出的自己推一下吧。 
13 
14 */
15 #include<stdio.h>
16 int euler(int n)
17 {
18     int ret=1;
19     for(int i=2;i*i<=n;i++)
20         if(n%i==0){
21             n/=i;ret*=i-1;
22             while(n%i==0){
23                 n/=i;ret*=i;
24             }
25         }
26     if(n>1) ret*=(n-1);
27     return ret;
28 }
29 int cul(int n,int m)
30 {
31     int sum=0;
32     for(int i=1;i*i<=n;i++){
33         if(n%i==0){ 
34             if(n/i>=m && i*i!=n)
35                 sum+=euler(i);
36             if(i>=m)
37                 sum+=euler(n/i);
38         } 
39     }
40     return sum;
41 }
42 int main(void)
43 {
44     int t,n,m;
45     scanf("%d",&t);
46     while(t--)
47     {
48         scanf("%d%d",&n,&m);
49         printf("%d
",cul(n,m));
50     }
51     return 0;
52 }