zoukankan      html  css  js  c++  java
  • poj3090 Visible Lattice Points [欧拉函数]

    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

    一个点(x,y)能被射中的充分必要条件是x,y互质,是不是很像欧拉函数?
    经过容斥可以得到,若用sum表示1...n的phi值,ans=sum*2+2-1=sum*2+1
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 using namespace std;
     5 #define dbg(x) cout<<#x<<" = "<<x<<endl
     6 
     7 const int maxn=1005;
     8 
     9 bool check[maxn];
    10 int prime[maxn],phi[maxn],n,dat[maxn],phisum[maxn];
    11 int ans=0,cnt=0,T;
    12 
    13 void getphi(){
    14     phi[1]=1;
    15     for(int i=2;i<=n;i++){
    16         if(!check[i]){
    17             prime[++cnt]=i;
    18             phi[i]=i-1;
    19         }
    20         for(int j=1;j<=cnt&&prime[j]*i<=n;j++){
    21             check[i*prime[j]]=1;
    22             if(i%prime[j])  phi[i*prime[j]]=phi[i]*(prime[j]-1);
    23             else{
    24                 phi[i*prime[j]]=phi[i]*prime[j];
    25                 break;
    26             }
    27         }
    28     }
    29     for(int i=1;i<=n;i++)  phisum[i]=phisum[i-1]+phi[i];
    30 }
    31 
    32 int main(){
    33     scanf("%d",&T);
    34     for(int i=1;i<=T;i++){
    35         scanf("%d",&dat[i]);
    36         n=max(dat[i],n);
    37     }
    38     getphi();
    39     for(int i=1;i<=T;i++){
    40         printf("%d %d %d
    ",i,dat[i],(phisum[dat[i]]<<1)+1);
    41     }
    42     return 0;
    43 }
  • 相关阅读:
    PHP编程资源
    JSP+Java编程资源
    Word、Excel办公书的资源下载
    听你说
    一些好看的渐变色(配色)网站推荐
    js判断数组中是否包含某个元素
    你才是你故事的作者
    vue-color 颜色选择器插件用法介绍
    vue-cli3 导入.md文件,vue中markdown文件的解析与渲染
    新版 animate.css 在vue中的正确使用
  • 原文地址:https://www.cnblogs.com/ZYBGMZL/p/7271650.html
Copyright © 2011-2022 走看看