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最新推荐文章于 2019-03-26 17:31:32 发布

Kblacksmith

最新推荐文章于 2019-03-26 17:31:32 发布

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分类专栏： acm

本文链接：https://blog.youkuaiyun.com/carryheart/article/details/52589400

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本文介绍了一个数学求和问题，即计算从1到n整数范围内1/k²的累加和，并通过编程实现求解。文章提供了一种预处理的方法来优化计算过程。

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Cure

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1137 Accepted Submission(s): 380

Problem Description

Given an integer

n , we only want to know the sum of

1/k2 where

k from

1 to

n .

Input

There are multiple cases.
For each test case, there is a single line, containing a single positive integer

n .
The input file is at most 1M.

Output

The required sum, rounded to the fifth digits after the decimal point.

Sample Input

Sample Output

Source

2016 ACM/ICPC Asia Regional Qingdao Online

打表会发现到后来答案会保持不变，所以打一张表记录变化的时候的情况。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
const int maxn=300000+100;
double dp[maxn];
char in[maxn];

void cal()
{
    double ans=0;
    int limit=200000;
    for (int i = 1 ; i <= limit; i++)
    {
        ans+=1.0/(1.0*i*i);
        dp[i]=ans;
    }
}

int main()
{
    int pos ;
    cal();
    while (scanf("%s",in)!=EOF)
    {
        int dis = strlen(in);
        if(dis > 6)
            pos = 200000;
        else
        {
            pos = 0;
            for(int i = 0; i < dis; i++)
            {
                pos = pos*10 + (in[i]-'0');
            }
        }
        if(pos > 200000)
           pos = 200000;
        printf("%.5f\n",dp[pos]);
    }
    return 0;
}