Consecutive sum II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s):
1219

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

3

0

1

2

Sample Output

0 1

1 8

8 27

找规律

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