PAT 甲级 1105 Spiral Matrix

https://pintia.cn/problem-sets/994805342720868352/problems/994805363117768704

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, where m and n satisfy the following: m×n must be equal to N; m≥n; and m−n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 10​4​​. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 93
42 37 81
53 20 76
58 60 76

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
int num[maxn];
int ans[1010][1010];

int main() {
    scanf("%d", &N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &num[i]);
    sort(num, num + N);
    for(int i = 0; i < N / 2; i ++)
        swap(num[i], num[N - i - 1]);

    int line, row;
    for(row = sqrt((double)N); row >= 1; row --) {
        if(N % row == 0) {
            line = N / row;
            break;
        }
    }

    int up = 0, down = line - 1, left = 0, right = row - 1, cnt = 0;
    while(true) {
        for(int j = left; j <= right; j ++) ans[up][j] = num[cnt ++];
        if(++ up > down) break;
        for(int i = up; i <= down; i ++) ans[i][right] = num[cnt ++];
        if(-- right < left) break;
        for(int j = right; j >= left; j --) ans[down][j] = num[cnt ++];
        if(-- down < up) break;
        for(int i = down; i >= up; i --) ans[i][left] = num[cnt ++];
        if(++ left > right) break;
    }

    for(int i = 0; i < line; i ++) {
        for(int j = 0; j < row; j ++) {
            printf("%d", ans[i][j]);
            printf("%s", j != row - 1 ? " " : "\n");
        }
    }
    return 0;
}

　　中午好！

FH

标签：PAT,Matrix,int,num,++,1105,line,left,row
来源： https://www.cnblogs.com/zlrrrr/p/10355978.html