zoukankan      html  css  js  c++  java
  • LightOJ 1245 数学

    Harmonic Number (II)

    Description

    I was trying to solve problem '1234 - Harmonic Number', I wrote the following code

    long long H( int n ) {
        long long res = 0;
        for( int i = 1; i <= n; i++ )
            res = res + n / i;
        return res;
    }

    Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.

    Input

    Input starts with an integer T (≤ 1000), denoting the number of test cases.

    Each case starts with a line containing an integer n (1 ≤ n < 231).

    Output

    For each case, print the case number and H(n) calculated by the code.

    Sample Input

    11

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    2147483647

    Sample Output

    Case 1: 1

    Case 2: 3

    Case 3: 5

    Case 4: 8

    Case 5: 10

    Case 6: 14

    Case 7: 16

    Case 8: 20

    Case 9: 23

    Case 10: 27

    Case 11: 46475828386

    思路:类似分块的思想最多有2sqrt(n)个数;

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define esp 0.00000000001
    #define pi 4*atan(1)
    const int N=1e3+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
    int main()
    {
        int T,cas=1;
        scanf("%d",&T);
        while(T--)
        {
            ll x;
            scanf("%lld",&x);
            ll ans=0;
            for(ll i=1;i*i<=x;i++)
            {
                if(i!=x/i)
                ans+=x/i;
                ll len=(x/i)-x/(i+1);
                ans+=i*len;
            }
            printf("Case %d: %lld
    ",cas++,ans);
        }
        return 0;
    }
  • 相关阅读:
    人件阅读笔记之三
    明日计划:团队开发Fooks第十天
    明日计划:团队开发Fooks第九天
    明日计划:团队开发Fooks第八天
    明日计划:团队开发Fooks第七天
    明日计划:团队开发Fooks第六天
    优先队列
    KMP
    django-中间件
    Ajax--参数,csrf跨站请求伪造,serialize(),上传文件formdata
  • 原文地址:https://www.cnblogs.com/jhz033/p/5785270.html
Copyright © 2011-2022 走看看