CodeForces 1353 C. Board Moves 动态规划

1. 题目描述

1.1. Limit

Time Limit: 1 second

Memory Limit: 256 megabytes

1.2. Problem Description

You are given a board of size $n \times n$, 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. $n^2-1$ cells should contain $0$ figures and one cell should contain $n^2$ figures).

You have to answer $t$ independent test cases.

1.3. Input

The first line of the input contains one integer $t$ ( $1 \le t \le 200$) — the number of test cases. Then $t$ test cases follow.

The only line of the test case contains one integer $n$ ( $1 \le n < 5 \cdot 10^5$) — 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 $5 \cdot 10^5$ ( $\sum n \le 5 \cdot 10^5$).

1.4. Onput

For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.

1.5. Sample Input

3
1
5
499993

1.6. Sample Onput

0
40
41664916690999888

1.7. Source

CodeForces 1353 C. Board Moves

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2. 解读

通过观察建立递推方程，从矩阵中心向外，第 $i$ 层有 $8i$ 个元素，需要移动 $i$ 次才能到达矩阵中心。

$f(i) = f(i-1) + 8 \times (i-1)^2$

将 $i$ 换为矩阵维度 $x$， $i = 2x-1$。

$f(2x-1) = f(2(x-1)-1) + 8 \times (2(x-1)-1)^2$

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$\text{Table 1.样例数据表}$

$x$	$2x-1$	结果	差值
1	1	0
2	3	8	8
3	5	40	32
4	7	114	72
5	9	506	392
6	11	1164	648

3. 代码

#include <algorithm>
#include <iostream>
#include <string.h>
const long long num = 5 * 1e5 + 1;
using namespace std;

long long list[num];

void calculate(long long n)
{
    list[1] = 0;
    // 递推方程
    // 这里的 i 一定要是 long long ，不然会溢出，猜测是(i - 1) * (i - 1)用了i的类型进行存储
    for (long long i = 2; i < n; i++) {
        list[i] = list[i - 1] + 8 * (i - 1) * (i - 1);
    }
}

int main()
{
    // test case
    long long t;
    long long n;
    // 计算
    calculate((num + 1) / 2);
    // test case
    scanf("%lld", &t);
    // for each test case
    while (t--) {
        // 输入
        scanf("%lld", &n);
        // 输出
        printf("%lld\n", list[(n + 1) / 2]);
    }
}

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