Consecutive sum II
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2237 Accepted Submission(s): 1093
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
题目大意：递推，输入N，输出为N*N*N (N+1)*(N+1)*(N+1)
#include<iostream>
using namespace std;
int main()
{
__int64 a;
int T;
cin >> T;
while(T--)
{
cin >> a;
cout << a*a*a << " " <<(a+1)*(a+1)*(a+1) << endl;
}
return 0;
}
HDU1977 Consecutive sum II【水题】
原文：http://blog.csdn.net/lianai911/article/details/42304393
