Problem Description
You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2.
Input
In each case, there is an odd positive integer n.
Output
Print the sum. Make sure the sum will not exceed 2^31-1
Sample Input
3
Sample Output
10
Calculate the formula
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2条回答 默认 最新
- 2017-06-03 16:17回答 1 已采纳 http://blog.csdn.net/qq_26525215/article/details/51319577
- 2017-10-07 14:20回答 1 已采纳 http://blog.csdn.net/chinaczy/article/details/4891859
- 2017-02-22 14:53回答 2 已采纳 http://download.csdn.net/search?keywords=molecular
- 2014-11-10 21:17weixin_30813225的博客 You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2. Input In each case, there is an odd positive integer n. Output Print the sum. Make sure the sum ...
- 2018-04-08 22:58weixin_34074740的博客 You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2. InputIn each case, there is an odd positive integer n.OutputPrint the sum. Make sure the sum will not exceed 2^31-1Sample ...
- 2014-04-27 03:54宅的一逼的博客 You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2. Input In each case, there is an odd positive integer n. Output Print the sum. Make sure the ...
- 2015-01-29 22:13ITCharge的博客 Calculate the formula Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7441 Accepted Submission(s): 2284 Problem Description ...
- 2017-12-01 17:50逸川先生的博客 很有意思，又是一个java过不了的题目。 之前遇到过因为输入输出慢而A不了的，但可以使用java的StreamTokenizer和PrintWriter来解决。 而这个题目，完全就不是这个输入输出的问题问题，根本没人用java A过。...
- 2018-09-21 13:46weixin_30549175的博客 Problem Description You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2. Input In each case, there is an odd positive integ...
- 2018-03-30 20:46FancyKing的博客 HDU-2139 Calculate the formula HDU-2139 Calculate the formula 题目描述 思路 代码 题目描述 You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2. Input In each ...
- 2014-11-10 09:42AC_Gibson的博客 纯数学题 #include #include using namespace std; int main() { long long n; while(cin>>n) printf("%lld\n",n*(n+1)*(n+2)/6); return 0; }
- 2016-01-30 20:48星空永恒&&卡利达的博客 #include __int64 num[100001]; int main() { int i,n; num[1]=1; num[3]=10; for(i=5;i;i=i+2) num[i]=num[i-2]+i*i; while(scanf("%d",&n)!=EOF) { printf("%I64d\n",num[n]); } ...}
- 2015-12-18 14:36黄小鱼ZZZ的博客 #include __int64 shu[2600]; int main() { int i; shu[1]=1; for(i=3;i;i+=2) shu[i]=shu[i-2]+i*i;... while(~scanf("%d",&n)) 这次我可以自豪的说我一次过了，足以说明我现在是有多水。...这么
- 2015-08-07 19:22Tth_1993的博客 import java.util.Scanner; public class p2139 { public static void main(String[] args) { Scanner sc=new Scanner(System.in);... //一开始测试数据知道当n大于2344的时候刚好溢出 2^31-1 ...
- 2016-05-05 01:57谙忆的博客 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 看到这个时间，我懵逼了。。。 果然，Java就是打表，都不能AC，因为Java的输入是流，需要...You just need to calculate the su
- 2017-11-27 17:12Whale-Song的博客 HDACM 2139由于Java的System.out.println(sum[n]);函数是 加了锁的（synchronized） 打印速度慢所以A不了import java.util.Scanner;public class Main{ public static void main(String[] args) { ...
- 2011-08-29 20:19weixin_30839881的博客 Calculate the formula Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3836Accepted Submission(s): 1126 Problem Description You just need ...
- 2012-10-15 13:10weixin_30613727的博客 点击打开链接 找规律，不能直接算，否则超时 n=(n+2)*(n+1)*n/6; #include<stdio.h> int main() { __int64 n; while(scanf("%I64d",&n)!=EOF) { n=(n+2)*(n+1)*n/6;... ret...
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