zoukankan      html  css  js  c++  java
  • [Usaco2008 Dec]Patting Heads

    It's Bessie's birthday and time for party games! Bessie has instructed the N (1 <= N <= 100,000) cows conveniently numbered 1..N to sit in a circle (so that cow i [except at the ends] sits next to cows i-1 and i+1; cow N sits next to cow 1). Meanwhile, Farmer John fills a barrel with one billion slips of paper, each containing some integer in the range 1..1,000,000. Each cow i then draws a number A_i (1 <= A_i <= 1,000,000) (which is not necessarily unique, of course) from the giant barrel. Taking turns, each cow i then takes a walk around the circle and pats the heads of all other cows j such that her number A_i is exactly divisible by cow j's number A_j; she then sits again back in her original position. The cows would like you to help them determine, for each cow, the number of other cows she should pat.


    题目大意就是
    给出n个数,然后对每个数,求出其他的数有几个是它的约数


    数的范围是到100W
    然后n的范围是10W


    不能直接n^2暴力求

    不过可以使用一种类似于筛法的方法。

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    int a[1111111];
    int num[1111111];
    int n, nt;
    int x[111111];
    int t[111111];
    void ready()
    {
        for(int i = 0; i < nt; i++)
            for(int j = x[i]; j <= 1000000; j += x[i])
                a[j] += num[x[i]];
    }
    int main()
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &x[i]), t[i] = x[i], num[x[i]]++;
        sort(x, x + n);
        nt = unique(x, x + n) - x;
        ready();
        for(int i = 0; i < n; i++) printf("%d
    ", a[t[i]] - 1);
        return 0;
    }


  • 相关阅读:
    if elseif else
    java编程思想第四版中net.mindview.util包
    eclipse git插件配置
    php面试常用算法
    数据库字段类型中char和Varchar区别
    MySQL的数据库引擎的类型及区别
    windows系统中eclipse C c++开发环境的搭建
    launch failed.Binary not found in Linux/Ubuntu解决方案
    技术团队的情绪与效率
    如何有效使用Project(2)——进度计划的执行与监控
  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3177629.html
Copyright © 2011-2022 走看看