Codeforces Round 871 (Div. 4)——G题讲解-Toy模板网

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Hello, 大家好哇！本初中生蒟蒻讲解一下G. Hits Different!
Codeforces Round 871 (Div. 4)——G题讲解
上绿名喽！

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G. Hits Different

题目描述

In a carnival game, there is a huge pyramid of cans with 2023 rows, numbered in a regular pattern as shown.

Codeforces Round 871 (Div. 4)——G题讲解

If can $9^2$ is hit initially, then all cans colored red in the picture above would fall.

You throw a ball at the pyramid, and it hits a single can with number $n^2$ . This causes all cans that are stacked on top of this can to fall (that is, can $n^2$ falls, then the cans directly above $n^2$ fall, then the cans directly above those cans, and so on). For example, the picture above shows the cans that would fall if can $9^2$ is hit.

What is the sum of the numbers on all cans that fall? Recall that $n^2=n×n$ .

Input

The first line contains an integer $t (1 \leq t \leq 1000)$ — the number of test cases.

The only line of each test case contains a single integer $n (1≤n≤10^6)$ — it means that the can you hit has label $n^2$ .

Output

For each test case, output a single integer — the sum of the numbers on all cans that fall.

Please note, that the answer for some test cases won’t fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language (like long long for C++). For all valid inputs, the answer will always fit into 64-bit integer type.

Example

Codeforces Round 871 (Div. 4)——G题讲解
Note
The first test case is pictured in the statement. The sum of the numbers that fall is
$1^2+2^2+3^2+5^2+6^2+9^2=1+4+9+25+36+81=156.$
In the second test case, only the can labeled $1^2$ falls, so the answer is $1^2=1$ .
In the third test case, the cans labeled $1^2$ and $2^2$ fall, so the answer is $1^2+2^2=1+4=5$ .
In the fourth test case, the cans labeled $1^2$ and $3^2$ fall, so the answer is $1^2+3^2=1+9=10$ .
In the fifth test case, the cans labeled $1^2, 2^2$ , and $4^2$ fall, so the answer is $1^2+2^2+4^2=1+4+16=21.$

思路

本题可以想到动态规划，我们可以这样考虑：
Codeforces Round 871 (Div. 4)——G题讲解
假设砸 $14^2$ 这个点，我们看一下怎么转移：

红色的是 $14^2$ 左边这个点 $7^2$ 这个点所能波及到的点

绿色的是 $14^2$ 右边的点 $8^2$ 所能波及到的点

然后我们看一下他们重复了那些点：
Codeforces Round 871 (Div. 4)——G题讲解
其实就是 $14^2$ 上一行的上一行的左边的点 $4^2$ 所能波及到的点

故此，砸 $14^2$ 的值应该是砸 $7^2$ 的值+砸 $8^2$ 的值-砸 $4^2$ 的值+ $14^2$

可以推导出状态转移方程：（ $d p [i] [j]$ 表示砸图中第i行第j列的点的答案）
$d p [i] [j] = d p [i - 1] [j] + d p [i - 1] [j - 1] - d p [i - 2] [j - 1] + n u m [i] [j]$

其中， $n u m [i] [j]$ 表示图中第i行第j列的数字

本题就是求 $d p [1 - 2023] [1 - 2023]$ 的值
时间复杂度 $O(n^2)$

题中是输入的 $n$ 我们找到他是第几行，第几列即可！文章来源地址https://www.toymoban.com/news/detail-435971.html

代码

#include <iostream>
#define int long long

using namespace std;

const int N = 2024;

int dt;
int n;
int dp[N][N];
int num[N][N];
int sum[N];

signed main()
{
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);
	
	int t = 1;
	for (int i = 1; i <= 2023; i ++)
		for (int j = 1; j <= i; j ++)
			num[i][j] = t * t, t ++;
			
	sum[1] = 1;
	for (int i = 2; i <= 2023; i ++)
		sum[i] = sum[i - 1] + i - 1;
	
	dp[1][1] = 1;
	for (int i = 2; i <= 2023; i ++)
		for (int j = 1; j <= i; j ++)
			dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] - dp[i - 2][j - 1] + num[i][j];
	
	cin >> dt;
	
	while (dt --)
	{
		cin >> n;
		
		int h;
		for (int i = 1; i <= 2023; i ++)
			if (n < sum[i])
			{
				h = i - 1;
				break;
			}
		
		cout << dp[h][n - sum[h] + 1] << endl;
	}
	
	return 0;
}