Blue Bridge cup-shaped back to the questions basic exercises take a few Python and c ++

Questions basic exercises back-shaped access

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Resource constraints

Time limit: 1.0s Memory Limit: 512.0MB

Problem Description

　　Access is meander-shaped along the side access of the matrix, if the current direction has numerous desirable or take off, then 90 degrees left. A start left corner of the matrix, the downward direction.

Input Format

　　The first input line is a positive integer of not more than two of m 200, n, denotes the matrix of rows and columns. Next, each row of n row m is an integer, denotes the matrix.

Output Format

　　Only the output line, a total of mn number, taken as an input matrix meandering number of results obtained. Between the number separated by a space, end of the line do not have extra spaces.

Sample input

3 3
1 2 3
4 5 6
7 8 9

Sample Output

1 4 7 8 9 6 3 2 5

Sample input

3 2
1 2
3 4
5 6

Sample Output

1 3 5 6 4 2

 

Python version

m, n = map(int, input().split())
a = []
ans = []
for _ in range(m):
    a.append(list(map(int, input().split())))
while a:
    # down
    for i in range(len(a)):
        ans += [a[i].pop(0)] if a[i] else []
    # right
    ans += a.pop() if a else []
    # top
    for i in range(len(a) - 1, -1, -1):
        ans += [a[i].pop()] if a[i] else []
    # left
    ans += a.pop(0)[::-1] if a else []
print(*ans)

c ++ version

#include <iostream>
#include <memory.h>
using namespace std;
int main() {
    int m, n;
    cin >> m >> n;
    int a[201][201];
    memset(a, -1, sizeof(a));
    int i = 0, j = 0;
    for(i = 0; i < m; i++)
        for(j = 0; j < n; j++)
            cin >> a[i][j];
    i = 0, j = 0;
    int total = 0;
    while(total < m * n) {
        while(i <= m-1 && a[i][j] != -1) {//down
            cout << a[i][j] << " ";
            a[i][j] = -1;
            i++;
            total++;
        }
        i--;
        j++;
        while(j <= n-1 && a[i][j] != -1) {//right
            cout << a[i][j] << " ";
            a[i][j] = -1;
            j++;
            total++;
        }
        j--;
        i--;
        while(i >= 0 && a[i][j] != -1) {//up
            cout << a[i][j] << " ";
            a[i][j] = -1;
            i--;
            total++;
        }
        i++;
        j--;
        while(j >= 0 && a[i][j] != -1) {//left
            cout << a[i][j] << " ";
            a[i][j] = -1;
            j--;
            total++;
        }
        j++;
        i++;
    } 
    return 0;
}