PAT (Advanced Level) Practise 1105. Spiral Matrix (25) 蛇形填数

1105. Spiral Matrix (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ncolumns, 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 10⁴. 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

基本上就是nyoj上一道蛇形填数的原题。。。连N的范围都没给。。。不过从时间限制150ms推测肯定不大，可以用数组存

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 10000 + 10;
int a[maxn];

//按题目要求找m和n
void f(int N, int& m, int& n) {
	for (int i = 1; i <= N; i++) {
		if (N % i == 0) {
			if (i < N / i) continue;
			if (i - N / i < m - n) {
				m = i;
				n = N / i;
			}
		}
	}
}

bool cmp(int x, int y) {
	return x > y;
}

int MAP[1005][1005];

int main()
{
	int N, m, n;
	scanf("%d", &N);
	for (int i = 1; i <= N; i++) {
		scanf("%d", &a[i]);
	}
	m = 11111, n = 0;
	f(N, m, n);

	sort(a + 1, a + 1 + N, cmp);
	//m row n col
	memset(MAP, -1, sizeof(MAP));
	int i = 1, j = 1, cnt = 1;
	while (cnt <= N) {
		//right
		while (MAP[i][j] == -1 && j <= n) {
			MAP[i][j] = a[cnt++];
			j++;
		}
		j -= 1; //注意往回走一步，不然越界了
		i += 1; //注意往下走一步，不然重复了

		//down
		while (MAP[i][j] == -1 && i <= m) {
			MAP[i][j] = a[cnt++];
			i++;
		}
		i -= 1;
		j -= 1;

		//left
		while (MAP[i][j] == -1 && j >= 1) {
			MAP[i][j] = a[cnt++];
			j--;
		}
		j += 1;
		i -= 1;

		//up
		while (MAP[i][j] == -1 && i >= 1) {
			MAP[i][j] = a[cnt++];
			i--;
		}
		i += 1;
		j += 1;
	}

	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			if (j == 1)
				printf("%d", MAP[i][j]);
			else
				printf(" %d", MAP[i][j]);
		}
		puts("");
	}
	return 0;
}