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Consecutive sum II

时间限制：3000 ms  |  内存限制：65535 KB

难度：2

描述

Consecutive sum come again. 
1    = 0 + 1
2+3+4    = 1 + 8
5+6+7+8+9  = 8 + 27
…
You can see the consecutive sum can be representing like that. The nth line will have 2*n+1 consecutive numbers on the left, the first number on the right equal with the second number in last line, and the sum of left numbers equal with two number’s sum on the right.
Your task is that tell me the right numbers in the nth line.

输入

The first integer is T, and T lines will follow.
Each line will contain an integer N (0 <= N <= 20000).

输出

For each case, output the right numbers in the Nth line.
All answer in the range of signed 64-bits integer.

样例输入

3012

样例输出

0 11 88 27

上传者

ACM_吕德铭

import java.util.*;public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in); int n=in.nextInt(); for(int i=0;i<n;i++){ long m=in.nextLong(); long mm=m+1; long a1=m*m*m; long a2=mm*mm*mm; System.out.println(a1+" "+a2);  }}}