HDU 1977 Consecutive sum II

http://acm.hdu.edu.cn/showproblem.php?pid=1977

 

Consecutive sum II

Time Limit: 3000/1000 MS(Java/Others)    MemoryLimit: 32768/32768 K (Java/Others)
Total Submission(s):894    AcceptedSubmission(s): 453

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. Thenth line will have 2*n+1 consecutive numbers on the left, the firstnumber on the right equal with the second number in last line, andthe sum of left numbers equal with two number’s sum on theright.
Your task is that tell me the right numbers in the nthline.

 

 

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 12

 

 

Sample Output

0 1 1 8 827

 

 

Author

Wiskey

 

 

Source

2008杭电集训队选拔赛

 

 

Recommend

wangye

 

分析：找规律。

   我找的规律：每一行左边共n^2+1个数，每行第一个数为2n+1，这样通过等差数列求和，化简可以得到：这一行左边加起来等于2*i*i*i+3*i*i+3*i+1.

  用数组biao[2100000]储存该数，按照题中的输出要求，给biao[]赋值时：biao[i]=2*i*i*i+3*i*i+3*i+1-biao[i-1].给定n，输出biao[n-1]biao[n]\n即可。

  我自己找的规律其实非常麻烦，在网上看到给定n，m=n+1,输出n*n*n m*m*m\n即可。原来规律是这个。。完全没有发现啊(⊙o⊙)

 

代码如下：

（1）按我找的规律写的：

#include<stdio.h>
#include<string.h>
long long biao[2100010];
void fun()
{
 memset(biao,0,sizeof(biao));
 __int64 i;
 biao[0]=1;
 for(i=1;i<2100010;i++)
    biao[i]=2*i*i*i+3*i*i+3*i+1-biao[i-1];
}
int main()
{
 int T,n;
 fun();
 scanf("%d",&T);
 while(T--)
 {
  scanf("%d",&n);
  if(n==0) printf("0 1\n");
  else printf("%I64d%I64d\n",biao[n-1],biao[n]);
 }
 return 0;
}

（2）网上看到的 转载自：yzy杨子衍

#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;
}