HDU 5879-Cure(1/n^2之和)
2016-09-21 13:33
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Cure
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1274 Accepted Submission(s): 429
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
1
2
4
8
15
Sample Output
1.00000
1.25000
1.42361
1.52742
1.58044
Source
2016 ACM/ICPC Asia Regional Qingdao Online
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题目意思:
求1/1^2到1/n^2之和,保留5位小数点。
解题思路:
打表预处理,搜出前2000000,因为是保留5位小数点所以较大的数直接近似成(π^2)/6,即1.64493。#include<cstdio> #include<iostream> #include<vector> #include<cmath> #include<cstring> #include<queue> #include<algorithm> #include<set> #include<queue> #define INF 0xfffffff using namespace std; const int maxn=2000000; typedef long long ll; double a[maxn]; char s[maxn*10]; void f()//打表 { a[0]=a[1]=1; for(int i=2; i<=maxn; ++i) a[i]=1/pow(i,2)+a[i-1]; } int main() { f(); long long n; while(~scanf("%s",s)) { if(strlen(s)<=7) { sscanf(s,"%I64d",&n);//取最大I64d长度的字符串s转换存入长整型n if(n<=1000000) printf("%.5lf\n",a ); else printf("1.64493\n");//六分之π平方 } else printf("1.64493\n"); } return 0; }
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