UVa 769 - Magic of David Copperfield 解題報告（思維）

 Magic of David Copperfield 

The well-known magician David Copperfield loves to show the following trick: a square withN rows and N columns of different pictures appears on a TV screen. Let us number allthe pictures in the following order:

1	2		N
N*(N-1)+1	N*(N-1)+2		N*N

Each member of the audience is asked to put a finger on the upper left picture (i.e., picture numberone) and The Magic begins: the magician tells the audience to move the finger k₁ times through thepictures (each move is a shift of the finger to the adjacent picture up, down, left or right providedthat there is a picture to move to), then with a slight movement of his hand he removes some of thepictures with an exclamation ``You are not there!", and ... it is true - your finger is not pointing toany of the pictures removed. Then again, he tells the audience to make k<sub>2</sub> moves, and so on. At theend he removes all the pictures but one and smiling triumphantly declares, ``I've caught you"(applause).

Just now, David is trying to repeat this trick. Unfortunately, he had a hard day before, and youknow how hard to conjure with a headache. You have to write a program that will help David tomake his trick.

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Each test case consists of a single integer number N ().

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Your program should write the following lines with numbers to the output file:

where k_i is a number of moves the audience should make on the i-th turn ().All k_ishould be different (i.e. k_i <> k_j when i <> j). are the numbers of the picturesDavid should remove after the audience will make k_i moves (the number of the pictures removed isarbitrary, but each picture should be listed only once, and at least one picture should be removed oneach turn).

A description of the every next turn should begin with a new line. All numbers on each line shouldbe separated by one or more spaces. After e iterations, all pictures except one should be removed.

Sample input 

1

3

Sample Output 

8     4 6
13    9
10    7 1
7     8
11    3 5

    解題報告： 很有意思的一題。給定一n*n矩陣，其中爲1到n*n所有數字。求一方法，每次任意走（上下左右）k步，刪除一些數字；直到只剩下一個數字。

    剛開始，我是這樣想的：正方形矩陣可以這樣分類可以按照（i+j）的奇偶性分爲兩組，這樣每次走奇數步，刪除邊界的一個數，直到刪除到最後一個數就可以了。

    寫完後WA了。再看一遍題意，發現要求k不能相同。ok，改下k，每次不同，+2。運行一遍，發現n=100時k超了10000……

    問題就來了，每次刪除一個數是不夠的。再想一方案，仍然每次走奇數步，每次刪除右下角的一整斜行數，A了。

    其實就是一思維題，細心好好想想就很容易了。代碼如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iomanip>
using namespace std;
#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
typedef long long LL;
typedef unsigned long long ULL;
void work();
int main()
{
#ifdef ACM
    freopen("in.txt", "r", stdin);
#endif // ACM
    work();
}

/***************************************************/

void work()
{
    int T;
    scanf("%d", &T);

    fff(cas, 1, T)
    {
        int n;
        scanf("%d", &n);

        typedef vector<int> vint;
        map<int, vint> mint;

        fff(i, 1, n) fff(j, 1, n)
            mint[i+j].push_back((i-1)*n + j);

        int k = 199;
        dff(i, n+n, 3)
        {
            printf("%d", k+=2);
            for(vint::iterator it = mint[i].begin(); it != mint[i].end(); it++)
                printf(" %d", *it);
            puts("");
        }

        if(T-cas) puts("");
    }
}