Cure
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1136 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.
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
题意:输入一个值N,求1/(k*k)的和(k=1,2,...N)。
思路:注意点:1.这个数N可能很大,题意很隐蔽,要注意,得用字符串来记录才行。2.显然题目要求计算的这个p级数是收敛的,利用傅里叶级数可以间接的求得这个级数和为(pi)^2/6,但题目要求的精度只有5位,所以可以得知当N足够大时,这五位小数将会趋于一个固定的值。3.N较小时最好边算答案边用一个数组记录结果,这样当再次输入N的值小于前面的N值时,就不用再重新算一遍结果了。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<string> #include<cmath> using namespace std; const int N_MAX = 1000000; int num[10]; double mem[N_MAX+1];//mem数组最多能记录到N为N_MAX时的结果 int main() { string s; int i=1; double sum=0; while (cin>>s) { if(s.size()>7)cout << 1.64493 << endl;//由后面的程序计算可得输入的数大于7位,值就固定了 else { memset(num, 0, sizeof(num)); for (unsigned int i = 0;i <= s.size();i++) num[i] = s[i]-'0';//将数字每一位都存在数组里,数组存放从高位到低位 int N = 0;//当输入的数小于等于七位时,将其转化为整形 int k = s.size(), j = 0; while (k) { N += num[j++] * pow(10, k - 1); k--; } if (N <= N_MAX&&mem[N] > 0) { printf("%.5f ", mem[N]); }//输入的N值在数组范围且有记录时 else if (N > N_MAX)cout << 1.64493 << endl; else { while (i <= N) { double a = i; sum += (1/(a*a));/////////注意,不要强制转换,会inf!! mem[i] = sum;//边计算边用数组记录不同N情况下的级数和 i++; } printf("%.5f ", mem[N]); } } } return 0; }