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

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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:

1237 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 9342 37 8153 20 7658 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nvoid 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 colmemset(MAP, -1, sizeof(MAP));int i = 1, j = 1, cnt = 1;while (cnt <= N) {//rightwhile (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//downwhile (MAP[i][j] == -1 && i <= m) {MAP[i][j] = a[cnt++];i++;}i -= 1;j -= 1;//leftwhile (MAP[i][j] == -1 && j >= 1) {MAP[i][j] = a[cnt++];j--;}j += 1;i -= 1;//upwhile (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]);elseprintf(" %d", MAP[i][j]);}puts("");}return 0;}

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