杭电2588 GCD（欧拉函数+gcd的应用）

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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

31 110 210000 72

Sample Output

16260/*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;}

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