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有一个 nn 的矩阵#xff08;n 为奇数#xff09;#xff0c;每个矩阵单元有一个物品#xff0c;每次操作你可将一个单元里的一个物品移动到该单元周围的八个单元里#xff0c;问最后只有一个单元有物品的情况下#xff0c;最少要多少次操作#xff1…题意
有一个 n×n 的矩阵n 为奇数每个矩阵单元有一个物品每次操作你可将一个单元里的一个物品移动到该单元周围的八个单元里问最后只有一个单元有物品的情况下最少要多少次操作
题目
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,j1); (i,j−1); (i,j1); (i1,j−1); (i1,j); (i1,j1); 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≤n5⋅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
分析(1)
找规律。可以看到正方形的每一个“圈”到中心的最小距离都是一样的数一下就能发现规律。然后把整个正方形分成八个小三角形八条线或者直接按照圈来统计都行。
AC代码
#includestdio.h
#includestring.h
#includealgorithm
using namespace std;
typedef long long ll;
int t;
ll n,ans;
int main()
{scanf(%d,t);while(t--){ans0;scanf(%lld,n);for(ll i2; in/21; i)ans(i-2)*(i-1);ans*8;for(ll i1; in/21; i)ans(i-1)*8;printf(%lld\n,ans);}return 0;
}分析二
我们可以将所有物品最终都汇聚到矩阵的中心的单元格那么以此为中心他周围的第一圈的单元格的物品只需要一步就可以第二圈就需要两步以此类推。 那么我们可以预处理出每一圈的单元格的数量乘以他对应的步数再求和即可。
AC代码
#include bits/stdc.h
using namespace std;
const int N 5e5 10;
typedef long long ll;
int f[N];
int main()
{f[0] 0;for(int i 1; i N; i )f[i] f[i - 1] 8;int t;cin t;while(t --){ll n;cin n;n n * n - 1;ll ans 0;for(int i 1; ; i ){if(n f[i]){n - f[i];ans 1ll * i * f[i];}elsebreak;}cout ans endl;}return 0;
}
