PAT (Advanced Level)1105 Spiral Matrix (25分)_1105 spiral matrix测试点一-优快云博客

本文链接：https://blog.youkuaiyun.com/JasonRaySHD/article/details/104185399

本篇博客介绍了一道编程题目，要求将给定的正整数序列以非递减顺序填充到螺旋矩阵中。矩阵有m行n列，满足m×n=N，且m-n最小。输入包含一个测试案例，输出为符合螺旋顺序的矩阵。解题关键包括确定m和n的值，以及按顺时针方向填充矩阵。注意避免数组未全部赋值和潜在的死循环问题。

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1105 Spiral Matrix (25分)

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

关键：
1、找m、n：从1开始到N遍历使得m-n最小，并满足其他要求的m、n；
2、获取输出结果：用二位数组存储要输出的结果，并分四次为其赋值，顺序为顺时针，利用变量round来表示当前循环次数，并根据最开始输入的数据个数N决定是否退出最外层循环。

坑点：
1、找m、n：要从1开始一直遍历到N（可以不用考虑特殊情况），而不能为了快只遍历到sqrt（N）
2、内部的四个赋值循环一定要检测数组data中的元素是否已经全部赋值完毕。
3、如果遇到测试点超时，应该是陷入了死循环，自己看看程序哪里没对。

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

bool cmp(int a,int b){
	return a>b;
}

const int maxn = 10000;
int G[maxn][maxn];

int main(){
	int m,n,N;
	scanf("%d",&N);
	int data[N];
	for(int i=0;i<N;i++){
		scanf("%d",&data[i]);
	}
	sort(data,data+N,cmp);
	int min = 0x7FFFFFFF;
	for(int i=1;i<=N;i++){
		if(N%i == 0){
			m = N/i;
			if(i>=m && i-m<min){
				min = i-m;
				n = i;
			}
		}
	}
	m = N/n;
	int num = 0,round=0;
	while(num<N){
		for(int i=round;i<m-round && num<N;i++){
			G[round][i] = data[num++];
		}
		for(int i=round+1;i<n-round && num<N;i++){
			G[i][m-round-1] = data[num++];
		}
		for(int i=m-round-2;i>=round && num<N;i--){
			G[n-round-1][i] = data[num++];
		}
		
		for(int i=n-round-2;i>=round+1 && num<N;i--){
			G[i][round] = data[num++];
		}
		round++;
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			printf("%d",G[i][j]);
			if(j<m-1)printf(" ");
		}
		if(i<n-1) printf("\n");
	}
	
	
	return 0;
}