杭电2588 GCD（欧拉函数+gcd的应用）

GCD

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1080 Accepted Submission(s): 492



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
/*
x从1到N ,GCD(N,x)>=M的个数,用欧拉函数可以求出来1--N与N互质的个数，找到1--N中是N的约数且大于M的值x，1--N/x的欧拉函数值就是x--N之间比M大的个数，不太明白为什么不会重复，应该是因为x是N的约数吧
加油！！！
Time:2014-9-16 11:44
*/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAX=10000+10;
int N,M;
int num[MAX];
int Eular(int M){
	int ans=M;
	int sq=M;
	for(int i=2;i*i<=sq;i++){
		if(M%i==0){
			ans=ans/i*(i-1);
			while(M%i==0)M/=i; 
		}
	}
	if(M>1)ans=ans/M*(M-1);
	
	return ans;
}
void solve(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&N,&M);
		int i,k=0;
		for(i=1;i*i<N;i++){if(N%i==0){
				if(N/i>=M)
					num[k++]=i;
				if(i>=M)
					num[k++]=N/i;	
			} 
			
		}
		if(i*i==N&&i>=M){
		num[k++]=i;
		}
		int ans=0;
		for(int i=0;i<k;i++)
			ans+=Eular(num[i]);
			
		printf("%d\n",ans);
	}
}
int main(){
	solve();
return 0;
}