G. Hits Different

https://codeforces.com/problemset/problem/1829/G

dp：设求的值：n。设置数组val[n]为第n个的答案。那么有递推:val[n] = val[n-id[n]] + val[n-id[n]+1] - val[n-2*(id[n]-1)]且以上均存在的情况下，id[n]表示第几层
采用离线算法，避免无用功。
代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
#define IOS ios::sync_with_stdio(false), cin.tie(0) ,cout.tie(0)
using namespace std;
#define int unsigned long long
const int N = 1e6 + 10;
int val[N], id[N];

signed main()
{
	IOS;
	int t; cin >> t;
	int maxn = 0;
	queue<int>q;
	while (t--)
	{
		int n; cin >> n;
		maxn = max(maxn, n);
		q.push(n);
	}
	int idx = 1;
	int num = idx;
	for(int i=1;i<=maxn;i++)
	{
		if (num == 0) { idx++; num = idx; }//记录id，每层有num个，如果清零那么idx增加，然后num从头开始
		id[i] = idx;
		val[i] += i * i;
		if (id[i - id[i]] == id[i] - 1)val[i] += val[i - id[i]];
		if (id[i - id[i] + 1] == id[i] - 1)val[i] += val[i - id[i] + 1];
		if (i - (id[i] - 1) * 2 > 0 and id[i - (id[i] - 1) * 2] == id[i] - 2)val[i] -= val[i - (id[i] - 1) * 2];
		num--;
	}
	while (!q.empty())
	{
		cout << val[q.front()] << '\n';
		q.pop();
	}
	return 0;
}