zoukankan      html  css  js  c++  java
  • 牛客2018年第二次acm暑期培训----H-Diff-prime Pairs

     题目描述:

    链接:https://www.nowcoder.com/acm/contest/141/H
    来源:牛客网
     

    时间限制:C/C++ 1秒,其他语言2秒
    空间限制:C/C++ 262144K,其他语言524288K
    64bit IO Format: %lld

    题目描述

    Eddy has solved lots of problem involving calculating the number of coprime pairs within some range. This problem can be solved with inclusion-exclusion method. Eddy has implemented it lots of times. Someday, when he encounters another coprime pairs problem, he comes up with diff-prime pairs problem. diff-prime pairs problem is that given N, you need to find the number of pairs (i, j), where and are both prime and i ,j ≤ N. gcd(i, j) is the greatest common divisor of i and j. Prime is an integer greater than 1 and has only 2 positive divisors.

    Eddy tried to solve it with inclusion-exclusion method but failed. Please help Eddy to solve this problem.

    Note that pair (i1, j1) and pair (i2, j2) are considered different if i1 ≠ i2 or j1 ≠ j2.

    输入描述:

    Input has only one line containing a positive integer N.
    
    1 ≤ N ≤ 107

    输出描述:

    Output one line containing a non-negative integer indicating the number of diff-prime pairs (i,j) where i, j ≤ N

    示例1

    输入

    3

    输出

    2

    示例2

    输入

    5

    输出

    6
    #include<iostream>
    #include<cstring>
    #define ll long long
    using namespace std;
    const int MAXSIZE=1e7+10;
    int Mark[MAXSIZE<<2];
    int prime[MAXSIZE];
     
    ll sum(ll n){
        if(n%2==0)return (n+1)*(n>>1);
        else return ((n+1)>>1)*n;
    }
    int  Prime(int n){
        int index = 0;
        memset(Mark,0,sizeof(Mark));
        for(int i = 2; i <= n; i++)
        {
            if(Mark[i] == 0){
                prime[index++] = i;
            }
            for(int j = 0; j < index && prime[j] * i <= n; j++)
            {
                Mark[i * prime[j]] = 1;
                if(i % prime[j] == 0){
                    break;
                }
            }
        }
        return index;
    }
    int main(){
        int n,total,i=2;
        ll s=0;
        ios::sync_with_stdio(false);
        cin>>n;
        total=Prime(n);
        s+=sum((ll)total-1)*2;
        while(total>=2){
            if(prime[total-1]*i<=n){
                s+=sum((ll)total-1)*2;
                i++;
            }
            else
                total--;
        }
        cout<<s<<endl;
        return 0;
    }
  • 相关阅读:
    Win10安装.NetFamework3.5
    SAN和NAS的区别
    raid10模型比raid01模型的冗余度高
    Linux——查找占用磁盘体积最大的前10个文件
    Nginx——端口负载均衡
    oneinstack——证书更新
    SpringBoot——IDEA使用 Spring Initializer快速创建项目【四】
    Nginx——请求head被过滤
    Linux—— 记录所有登陆用户的历史操作记录
    Nginx——跨域造成的504问题
  • 原文地址:https://www.cnblogs.com/ke-yi-/p/10175837.html
Copyright © 2011-2022 走看看