GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1294 Accepted Submission(s): 583
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.
(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
题意:
计算1-N区间里有多少数和N的GCD是大于M的。
解题思路:
直接计算绝对超时,所以要想到采用一些定理来进行优化。
①我们先看两个数 N = a*b,X= a*d。因为gcd ( N , X ) = a 所以b,d这两个数互质。又因为d可以是任何一个小于b的数。所以d值数量的的多少就是b的欧拉函数值。
所以,我们可以枚举a,然后去求b,然后再求b的欧拉函数值。
②但是如果单纯这样全部枚举的话依旧会超时,所以我们要想一个办法去优化它。
我们可以折半枚举,这里的折半并不是二分的意思。
我们先看,我们枚举时,当i<sqrt(n),假设a=n / i, 当i>sqrt(n)之后 有b=n/i,我们观察到当n%i==0时,会出现一种情况,就是a*b==n。所以我们就可以只需要枚举sqrt(n)种情况,然后和它对应的情况就是 n/i。
我们这种枚举时间会快非常多。
AC代码:
#include <stdio.h> #include <math.h> #include <vector> #include <queue> #include <string> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> using namespace std; int euler(int n) { int res=n; for(int i=2;i*i<=n;i++){ if(n%i==0){ res=res/i*(i-1); while(n%i==0) n/=i; } } if(n>1) res-=res/n; return res; } int main() { int t; scanf("%d",&t); while(t--){ int n,m; scanf("%d%d",&n,&m); int ans=0; for(int i=1;i*i<=n;i++){ if(n%i==0){ if(i>=m)ans+=euler(n/i); //计算sqrt(n)左边的 if(n/i>=m&&i*i!=n) ans+=euler(i);//计算sqrt(n)右边的i*i==n时,在上个语句已经执行 } } printf("%d\n",ans); } return 0; }