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  • POJ-3090 Visible Lattice Points(欧拉函数)

    Visible Lattice Points
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 6826   Accepted: 4103

    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

     
    其实吧,和前前道题是一模一样的,然而我这个傻逼写了三次欧拉函数了,还是不能一遍写对QAQ
     1 #include <cstdio>
     2 #include <cmath>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <queue>
     6 #include <stack>
     7 #include <vector>
     8 #include <iostream>
     9 #include "algorithm"
    10 using namespace std;
    11 typedef long long LL;
    12 const int MAX=1005;
    13 int n,T;
    14 int phi[MAX];
    15 void euler(){
    16     int i,j;
    17     for (i=1;i<MAX;i++) phi[i]=i;
    18     for (i=2;i<MAX;i+=2) phi[i]/=2;
    19     for (i=3;i<MAX;i+=2)
    20      if (phi[i]==i)
    21       for (j=i;j<MAX;j+=i)
    22        phi[j]=phi[j]/i*(i-1);
    23 }
    24 int main(){
    25     freopen ("visible.in","r",stdin);
    26     freopen ("visible.out","w",stdout);
    27     int i,j,cas(0);
    28     scanf("%d",&T);
    29     euler();
    30     while (T--){
    31         LL ans(0);
    32         scanf("%d",&n);
    33         for (i=1;i<=n;i++)
    34          ans=ans+(LL)phi[i];
    35         printf("%d %d %lld
    ",++cas,n,ans*2+1);
    36     }
    37     return 0;
    38 }
    没意思,别看
    未来是什么样,未来会发生什么,谁也不知道。 但是我知道, 起码从今天开始努力, 肯定比从明天开始努力, 要快一天实现梦想。 千里之行,始于足下! ——《那年那兔那些事儿》
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  • 原文地址:https://www.cnblogs.com/keximeiruguo/p/6063552.html
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