zoukankan      html  css  js  c++  java
  • 算法复习——欧拉函数(poj3090)

    题目:

    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

    题解:

    欧拉函数模板题。

    心得:

    感觉欧拉函数稍微考得隐晦点的就是可视点问题了···嗯就这样

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cmath>
    #include<ctime>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cctype>
    using namespace std;
    const int N=1006;
    int phi[N],sum[N],n;
    void pre()
    {
      for(int i=1;i<N;i++)
        phi[i]=i;
      for(int i=2;i<N;i++)  
        if(phi[i]==i)
          for(int j=i;j<N;j+=i)
            phi[j]=phi[j]*(i-1)/i;
      for(int i=2;i<N;i++)
        sum[i]=phi[i]+sum[i-1];
    }
    int main()
    {  
      //freopen("a.in","r",stdin);
      pre();
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
      {
        int k;
        scanf("%d",&k);
        cout<<i<<" "<<k<<" "<<sum[k]*2+3<<endl;
      }
      return 0;
    }
  • 相关阅读:
    4.19Java.util.Arrays类
    4.19Java数组的拷贝
    Inverse matrix of 4x4 matrix
    自言自语
    病了两天
    当年3ds max盗版光碟上的广告
    头晕的厉害
    复习了一下STL容器的知识
    一个简单的能处理MIPMAP的类
    空间变换代码,相当简洁优美
  • 原文地址:https://www.cnblogs.com/AseanA/p/6663753.html
Copyright © 2011-2022 走看看