<h1 id="heading">题目</h1>
<blockquote>
<blockquote>
<h2 id="description">Description</h2>
Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
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.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
The required sum, rounded to the fifth digits after the decimal point.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
1 
2 
4 
8 
15
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
1.00000 
1.25000 
1.42361 
1.52742 
1.58044
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
求 <code>Σ1/n2</code> 
由于该式在 <code>n→∞</code> 时,结果为 <code>π&lt;sup&gt;2&lt;/sup&gt;/6</code> ( <code>≈1.64493</code> ) 
由于只需要输出 <code>5</code> 位小数,因此当 <code>n</code> 足够大时,只需要输出一个确定值( <code>1.64493</code> )即可
本地调试测试可知,当数据大于 <code>200000</code> 时,就可以放心输出 <code>1.64493</code> 了 
因此可以先打表计算 <code>200005</code> 内的值,当读入的数大于 <code>200005</code> 时,直接输出即可
由于 <code>n</code> 的大小不确定,因此应该高精度读入(注意前导 0 ) 
然后类似高精度的比较大小即可
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリ0隶�
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;
#include &lt;iomanip&gt;
#include &lt;iostream&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;string&gt;
#include &lt;vector&gt;
#include &lt;bitset&gt;
#include &lt;functional&gt;

using namespace std;

typedef long long LL;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

const int maxn = 200005;

LL n;
double ans[maxn];
char s[1000000];

char cmp[] = &#34;200000&#34;;

int read_string(char s[]) {
 char c;
 int i = 0;
 while(!((c = getchar()) &gt;= &#39;1&#39; &amp;&amp; c &lt;= &#39;9&#39;))
 if(c == EOF)
 return -1;
 while(c &gt;= &#39;0&#39;&amp;&amp;c &lt;= &#39;9&#39;) {
 s[i++] = c;
 c = getchar();
 }
 s[i] = &#39;\0&#39;;
 return i;
}

bool cmp_biger(char *a,char *b) {
 int len1 = strlen(a);
 int len2 = strlen(b);
 if(len1 &gt; len2)
 return true;
 else if(len1 == len2) {
 len1 = 0;
 while(a[len1] == b[len1] &amp;&amp; len1 &lt;= len2)
 len1++;
 if(len1 &lt;= len2 &amp;&amp; a[len1] &gt; b[len1])
 return true;
 else
 return false;
 } else
 return false;
}

bool Do() {
 if(read_string(s) == -1)
 return false;

 //cout &lt;&lt; cmp &lt;&lt; &#34; &#34; &lt;&lt; s &lt;&lt; endl &lt;&lt; &#34; &#34;;

 if(cmp_biger(cmp,s)) {
 int len = strlen(s);
 n = 0;
 for(int i = 0;i &lt; len;i++)
 n = n * 10 + s[i] - &#39;0&#39;;
 cout &lt;&lt; fixed &lt;&lt; setprecision(5) &lt;&lt; ans[n] &lt;&lt; endl;
 } else {
 cout &lt;&lt; &#34;1.64493&#34; &lt;&lt; endl;
 }
 return true;
}


int main() {
 cin.tie(0);
 cin.sync_with_stdio(false);


 ans[1] = 1.0;
 for(int i = 2;i &lt; maxn;i++)
 ans[i] = ans[i - 1] + 1 / (double)i / (double)i;


 while(Do());

 return 0;
}
</pre>

HDU 5879.Cure(2016 ACM/ICPC Asia Regional Qingdao Online)