GCD
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1320 Accepted Submission(s): 597
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
解题思路:
这道题一看到 2<=N<=1000000000,应该想到普通的做法肯定要超时,于是要想方法避免超时,“二分思想” 在这得以利用,左右开弓同时查找可以节省大量时间,并且许多合数都不必考虑,代码也从原来的TLE变成0ms.
#include<stdio.h>
int eular(int n)
{
int rt = 1,i;
for( i =2; i*i<=n; i++)
if(n%i==0)
{
n/=i;
rt=rt*(i-1);
while(n%i==0)
{
n/=i;
rt*=i;
}
}
if(n >1 ) rt =rt*(n-1);
return rt;
}
int main()
{
int t,n,m,x,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int ans = 0;
for(i = 1; i*i <= n ;i++)
{
if(n%i==0)
{
if(i>=m)
ans+=eular(n/i);
if(n/i>=m&&i*i!=n)
ans+=eular(i);
}
}
printf("%d\n",ans);
}
return 0;
}