HDU 2588 GCD 欧拉函数
2014-10-15 22:27
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GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1089 Accepted Submission(s): 498
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
/* HDOJ 2588 欧拉函数 给定N,M(2<=N<=1000000000, 1<=M<=N), 求1<=X<=N 且gcd(X,N)>=M的个数。 先找出N的约数x,并且gcd(x,N)>= M,结果为所有N/x的欧拉函数之和。 因为x是N的约数,所以gcd(x,N)=x>= M; 设y=N/x,y的欧拉函数为小于y且与y互质的数的个数。 设与y互质的的数为p1,p2,p3,…,p4, 则: x<x* pi<N, 满足1<=X<=N; 那么gcd(x* pi,N)=x>= M。pi是与y质数 */ #include<iostream> #include<stdio.h> using namespace std; int phi(int x) { int i,ans=x; for(i=2;i*i<=x;i++) { if(x%i==0) { ans-=ans/i; while(x%i==0) { x=x/i; } } } if(x>1) ans-=ans/x; return ans; } int solve(int n,int m) { int i,sum=0; for(i=1;i*i<=n;i++) { if(n%i!=0) continue; if(i>=m&&i*i!=n)//当i=nn时,n/i=nn,要避免求了两次欧拉函数 sum+=phi(n/i); if(n/i>=m) //对称的 sum+=phi(i); } return sum; } int main() { int n,m,t; //freopen("test.txt","r",stdin); scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); printf("%d\n",solve(n,m)); } return 0; }
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