zoukankan      html  css  js  c++  java
  • 数论

    Visible Lattice Points
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5636   Accepted: 3317

    Description

    A lattice point (xy) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (xy) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (xy) with 0 ≤ xy ≤ 5 with lines from the origin to the visible points.

    Write a program which, given a value for the size, N, computes the number of visible points (xy) with 0 ≤ xy ≤ N.

    Input

    The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.

    Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.

    Output

    For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.

    Sample Input

    4
    2
    4
    5
    231

    Sample Output

    1 2 5
    2 4 13
    3 5 21
    4 231 32549

    Source

     

     

    Mean: 

     在第一象限中,输入一个n,然后要你统计在(0<=x<=n,0<=y<=n)的范围内,有多少可视点。

    所谓的可视点,即:从(0,0)出发到达(x1,y1),中间未与任何整点相交的点。

    analyse:

     通过分析,我们会发现:只要x和y互质,那么(x,y)就是可视点。我们只要求得[0,0]~[x,y]内满足x和y互质的点(x,y)的个数,那么问题就可迎刃而解。欧拉函数就是用来解决小于n的数中有多少个数与n互质。

    Time complexity:O(n)

    Source code:

    // Memory   Time
    // 1347K     0MS
    // by : Snarl_jsb
    // 2014-09-12-22.35
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<vector>
    #include<queue>
    #include<stack>
    #include<map>
    #include<string>
    #include<climits>
    #include<cmath>
    #define N 1000010
    #define LL long long
    using namespace std;
    
    int gcd(int a,int b){
    	return b?gcd(b,a%b):a;
    }
    
    inline int lcm(int a,int b){
    	return a/gcd(a,b)*b;
    }
    
    int eular(int n)    ////求1..n-1中与n互质的数的个数
    {
    	int ret=1,i;
    	for (i=2;i*i<=n;i++)
    		if (n%i==0){
    			n/=i,ret*=i-1;
    			while (n%i==0)
    				n/=i,ret*=i;
    		}
    	if (n>1)
    		ret*=n-1;
    	return ret;
    }
    
    int main()
    {
    //    freopen("C:\Users\ASUS\Desktop\cin.cpp","r",stdin);
    //    freopen("C:\Users\ASUS\Desktop\cout.cpp","w",stdout);
        int t,Cas=1;
        cin>>t;
        while(t--)
        {
            int n;
            cin>>n;
            LL ans=0;
            for(int i=1;i<=n;i++)
            {
                ans+=eular(i);
            }
            printf("%d %d %d
    ",Cas++,n,ans*2+1);
        }
        return 0;
    }
     
    

      

  • 相关阅读:
    Html.Partial和Html.RenderPartial, Html.Action和Html.RenderAction的区别
    cygwin下git出现cabundle.crt相关错误的解决办法
    Orchard CMS前台页面为什么没有Edit链接?
    Entity Framework练习题
    分析Autofac如何实现Controller的Ioc(Inversion of Control)
    在Winform,Silvelight,WPF等程序中访问Asp.net MVC web api
    适合.net程序员的.gitignore文件
    如何处理Entity Framework中的DbUpdateConcurrencyException异常
    Asp.net MVC中repository和service的区别
    smplayer中使用srt字幕乱码问题
  • 原文地址:https://www.cnblogs.com/crazyacking/p/3969260.html
Copyright © 2011-2022 走看看