HDU1977 Consecutive sum II【数学计算+水题】

Consecutive sum II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3061    Accepted Submission(s): 1533

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

Author

Wiskey

Source

2008杭电集训队选拔赛

问题链接：HDU1977 Consecutive sum II。

题意简述：参见上文。

问题分析：（略）

程序说明：（略）

题记：说好的找规律，要找出其规律也许没那么容易，知道了也容易了。

AC的C语言程序如下：

/* HDU1977 Consecutive sum II */

#include <stdio.h>

int main(void)
{
    int t, n;

    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);

        printf("%lld %lld
", (long long) n * n * n, (long long) (n + 1) * (n + 1) * (n + 1));
    }

    return 0;
}