hdu 1977 Consecutive sum II

Consecutive sum II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2013    Accepted Submission(s): 975

Problem Description

Consecutive sum come again. Are you ready? Go ~~
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.

 

Input

The first integer is T, and T lines will follow.
Each line will contain an integer N (0 <= N <= 2100000).

 

Output

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

 

Sample Input

3012

 

Sample Output

0 11 88 27

 

序列：0 1 8 27 64 125 216    （除前两项外 后面数据有规律）

 

#include <stdio.h>int main (){__int64 n,t,m;scanf("%I64d",&t);while (t--){   scanf("%I64d",&n);   m=n+1;   printf("%I64d %I64d\n",n*n*n,m*m*m);}return 0;}