1021. There was a mountain in the past, and 1021 there was a mountain in the past

1021. There was a mountain in the past, and 1021 there was a mountain in the past
Description

In the past there was a mountain. The top view of the mountain was a n × n rectangle. () The lowest altitude was 1, and then the altitude rose along the ring.

Given the value of n, output the height chart of the mountain. Input Format

The input contains only one row, which is a positive integer n. Output Format

The output is the altitude chart of the mountain. Sample Input

4

Sample Output

     1     2     3     4    12    13    14     5    11    16    15     6    10     9     8     7

 

 

Shanghai Jiao Tong University acm library, address http://acm.sjtu.edu.cn/OnlineJudge/problem/1021

Google once, the existing practices on the Internet do not seem to be as good as mine, so I posted my practices ^_^.

 

The basic idea is to open up a n * n matrix, fill in a number indicating the height, and then output this matrix.

The question is how to fill in the correct number.

 

Take n = 5 as an example. Observe the number in the outermost circle and divide it into four parts,

Obviously, each part can be filled with a loop. Normally, the pseudocode that fills the entire circle is:

for i = 1 .. N-1
    map[1][i] = i
    map[i][N] = N - 1 + i
    map[N][N - i] = 2 * (N - 1) + i
    map[N - i][1]= 3 * (N - 1) + i

For non-outermost loops, record the offset of the starting filling position (including the X axis offset, Y axis offset, and fill digit offset.

Finally, for an odd side length matrix, you must fill in the maximum height (that is, the square of the side length) in the center of the matrix.

#include <iostream>#include <iomanip>#include <string.h>using namespace std;int main(){int i, j;int n;int *map;cin >> n;map = new int[n * n];memset(map, 0, sizeof(map));int h_offset = 1;int x_offset = 0;int y_offset = 0;for (int len = n - 1; len > 0; len-= 2){//a loop fills a circle, from outer to innerfor (i = 0; i < len; ++i){map[x_offset * n + (y_offset + i)] = h_offset + i;map[(x_offset + i) * n +  (n - 1 - y_offset)] = h_offset + len + i;map[(n - 1 - x_offset) * n + (n - 1 - y_offset - i)] = h_offset + 2 * len + i;map[(n - 1 - x_offset - i) * n + (y_offset)] = h_offset + 3 * len + i;}h_offset += 4 * len;x_offset ++;y_offset ++;}if(n % 2 == 1)map[n * n / 2] = n * n;for (i = 0; i < n; ++i){for (j = 0; j < n; ++j){cout << setw(6) << map[i * n + j];}cout << endl;}delete []map;return 0;}

Sixth-grade primary school mathematics Haidian xiaoyuan monkey peach

1/3 + 1/5 + 1/6 = 7/10
21/7*10 = 30 (count)
30*1/3 = 10; 30*1/5 = 6; 30*1/6 = 5 (count)
The old man and the old man have 10, 6, and 5 respectively.
(If you do not understand it, you can calculate it as: 1/3: 1/5: 1/6 =, and then get the answer (proportional relationship ))