1160 snake matrix, 1160 snake

1160 snake matrix, 1160 snake
1160 snake Matrix

 

Time Limit: 1 s space limit: 128000 KB title level: Silver QuestionDescription Description

James plays a digital game and obtains an n-row and n-column Number Matrix (where n is an odd number not greater than 100). The number filling method is as follows: in the center of the matrix, start from 1 and circle it in a counter-clockwise direction, expanding circle by circle until n rows and n columns are filled with numbers. Please output the square matrix of n rows and the sum of its diagonal numbers.

Input description Input Description

N (n rows and n columns)

Output description Output Description

A matrix composed of n + 1 rows and n actions. The last row is the sum of diagonal numbers.

Sample Input Sample Input

3

Sample output Sample Output

5 4 3
6 1 2
7 8 9
25

1 # include <iostream> 2 using namespace std; 3 int now = 1; 4 int a [101] [101]; 5 int s = 0; 6 int fx = 1; // 1 right 2 left 3 up 4 down 7 int tot = 1; 8 int ans = 0; 9 int main () 10 {11 int n; 12 cin> n; 13 s = n/2 + 1; 14 int I = s; 15 int j = s; 16 a [I] [j] = now; 17 now ++; 18 while (tot! = N * n) 19 {20 if (fx = 1 & j-I = 1) 21 {22 fx = 3; 23 // a [I] [j] = now; 24 // now ++; 25 // tot ++; 26} 27 if (fx = 2 & I = j) 28 {29 fx = 4; 30 // a [I] [j] = now; 31 // now ++; 32 // tot ++; 33} 34 if (fx = 3 & (I + j = n + 1 )) 35 {36 fx = 2; 37 // a [I] [j] = now; 38 // now ++; 39 // tot ++; 40} 41 if (fx = 4 & (I + j = n + 1) 42 {43 fx = 1; 44 // a [I] [j] = now; 45 // now ++; 46 // tot ++; 47} 48 if (fx = 1) // 1 right 2 left 3 Top 4 bottom 49 {50 j ++; 51 a [I] [j] = now; 52 now ++; 53 tot ++; 54} 55 if (fx = 2) // 1 right 2 left 3 Top 4 bottom 56 {57 j --; 58 a [I] [j] = now; 59 now ++; 60 tot ++; 61} 62 if (fx = 3) // 1 right 2 left 3 Top 4 lower 63 {64 I --; 65 a [I] [j] = now; 66 now ++; 67 tot ++; 68} 69 if (fx = 4) // 1 right 2 left 3 Top 4 bottom 70 {71 I ++; 72 a [I] [j] = now; 73 now ++; 74 tot ++; 75} 76} 77 for (int I = 1; I <= n; I ++) 78 {79 for (int j = 1; j <= n; j ++) 80 {81 cout <a [I] [j] <"; 82 if (I + j = n + 1) | (I = j )) 83 ans = ans + a [I] [j]; 84} 85 cout <endl; 86} 87 cout <ans; 88 return 0; 89}