题目描述
You are given a board of size n×n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i,j) you can move the figure to cells:
(i−1,j−1);
(i−1,j);
(i−1,j+1);
(i,j−1);
(i,j+1);
(i+1,j−1);
(i+1,j);
(i+1,j+1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n2−1 cells should contain 0 figures and one cell should contain n2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1≤n<5⋅105) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5⋅105 (∑n≤5⋅105).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
input
3
1
5
499993
output
0
40
41664916690999888
题目大意
给你一个n*n的方阵，n肯定为奇数。每个方阵上有一个图形，问:最少需要多少步，把所有的图形集中在一个图形上。可以斜着走
题目分析
n=3时
1 1 1
1 0 1
1 1 1
答案为：f[i]=8 * 1=8
n=5时
2 2 2 2 2
2 1 1 1 2
2 1 0 1 2
2 1 1 1 2
2 2 2 2 2
答案为：f[i]=+f[i-2]+8 * 2 * 2=40
…
由此可以推出：f[i]=f[i-2]+(i/2)*(i/2)*8
这样就得到了递推式。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=5e5+5;
LL f[N];
void init() //预处理出答案
{
for(int i=3;i<N;i+=2)
f[i]=f[i-2]+(LL)(i/2)*(i/2)*8; //(i/2)*(i/2)*8可能会爆int
}
int main()
{
init();
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
cout<<f[n]<<endl;
}
return 0;
}