题目描述：

题目链接

Given an integer n, we only want to know the sum of 1/（k * k） 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

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2
3
4
5
1
2
4
8
15

Sample Output

1
2
3
4
5
1.00000
1.25000
1.42361
1.52742
1.58044

题解：

需要注意的问题：

  1. 计算sum的时候不要1.0 / (i * i)这样计算， 要1.0 / i / i这样计算。
  2. 题目中没说n额具体范围， 只说了n >= 0 那就认为n无限大， 用字符串存储， 也同时意味着有规律可寻。

ac代码：

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//#include <bits/stdc++.h>
#include <iostream>
#include <stack>
#include "algorithm"
#include "cstdio"
#include "queue"
#include "set"
#include "cstring"
#include "string"
#include "map"
#include "vector"
#include "math.h"
#include "utility"   // pair头文件
#define esp 1e-6
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int N = 1e6 + 5;
const int M = 1e9 + 5;

double a[N];
int main(){
    a[1] = 1;
    for (int i = 2; i < N; i++){
        a[i] = a[i - 1] + double(1.0 / i / i);
    }
    string n;
    while(cin >> n){
        if (n.size() >= 7)
            printf("%.5f\n",a[N - 1]);
        else{
            int m = atoi(n.c_str());// 将字符串转化为整型
            printf("%.5f\n",a[m]);
        }
    }
}
1
恰似你一低头的温柔，娇弱水莲花不胜寒风的娇羞， 我的心为你悸动不休。  --mingfuyan