1105 Spiral Matrix
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 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
知识点:简单模拟;vector的使用
思路:
矩阵应该利用vector的数组来构建;因为如果是形成比较方正的矩阵,长和宽最大是100左右;但如果是质数,矩阵就会变成长条形长在10000以内,这样利用普通二维数组数组会超限;利用了vector容器可以.resize()的特点。
单列的输出特殊处理
1 #include <iostream> 2 #include <string> 3 #include <algorithm> 4 #include <cmath> 5 #include <vector> 6 using namespace std; 7 const int maxn = 1000; 8 9 vector<int> matrix[maxn]; 10 11 bool cmp(int a,int b){ 12 return a>b; 13 } 14 15 int getLong(int n){ 16 int i; 17 for(i=sqrt(n)+0.9;i<=n && n%i!=0;i++); 18 return i; 19 } 20 21 int main(){ 22 vector<int> list; 23 int n,tmp; cin >> n; 24 for(int i=0;i<n;i++){ 25 scanf("%d",&tmp); 26 list.push_back(tmp); 27 } 28 sort(list.begin(), list.end(), cmp); 29 int l = getLong(n); 30 int w = n/l; 31 for(int i=1;i<=l;i++){ 32 matrix[i].resize(w+1); 33 } 34 int x=1, y=1, ptr=0; 35 36 int base=0; 37 while(ptr<list.size()){ 38 if(w==1){ 39 for(;y<=l+base;y++){ 40 matrix[y][x]=list[ptr++]; 41 } 42 break; 43 } 44 for(; x<=w+base; x++){ 45 46 matrix[y][x]=list[ptr++]; 47 } 48 49 x--; y++; 50 for(; y<l+base; y++){ 51 matrix[y][x]=list[ptr++]; 52 } 53 for(; x>=1+base; x--){ 54 matrix[y][x]=list[ptr++]; 55 } 56 x++; y--; 57 for(; y>1+base; y--){ 58 matrix[y][x]=list[ptr++]; 59 } 60 x++; y++; 61 w-=2; l-=2; 62 base++; 63 //printf("%d %d w=%d l=%d",x,y,w,l); 64 65 } 66 for(int i=1;i<=getLong(n);i++){ 67 for(int j=1;j<=(n/getLong(n));j++){ 68 if(j!=1) printf(" "); 69 printf("%d",matrix[i][j]); 70 } 71 printf("\n"); 72 } 73 }
posted on 2018-11-09 10:52 iojafekniewg 阅读(212) 评论(0) 编辑 收藏 举报