Article seen on tv canada Categories greedy algorithm (24) dynamic programming (34) Search (14) tre

Prizes seen on tv canada for resource sharing Bowen connotation June recommended articles summarized Microsoft Azure Programming Contest questions concern hero CSDN community micro-letters, a lot of community welfare Q & A: Ye Jinfeng game engine architecture seen on tv canada describes the elements of a square matrix of order n is 1, 2,. .., and equal to n ^ 2, that each row, each column and the two elements on the diagonal, so that the square is called Cube. n is odd we have one kind of construction method, called "the upper right," such as when n = 3,5,7 given cube below 3 8 1 6,357,492,517,241,815,235 7,141,646,132,022,101,219 21 3,111,825,297,303,948,110 19 28 38 47 7 9,182,729,466,817,263,537 5,141,625,343,645,131,524 334,244,421,233,241 4,331,222,314,049,211 20 lines in the middle of the first number is always 1, the last line is the middle number n ^ 2, his right is 2, from the three cube You can see "at the top right" is what is intended.
# Include <stdio.h> # include <string.h> int main () {int i, j, n, m, a [20] [20], vis [20] [20], ans; scanf ("% d ", & m); while (m -) {scanf ("% d ", & n); memset (a, 0, sizeof (a)); memset (vis, 0, sizeof (vis)); i = 1 ;/ * 1 is always the first line * / j = (n +1) / 2 ;/ * intermediate that number * / ans = 1 ;/ * record from 1 * / while (ans <= n * n) / * 1 -> n ^ 2 * / {a [i] [j] = ans ;/ * record the current position value * / vis [i] [j] = 1; / * Record the current position to update * / if (i == 1 && j = n!) / * the first line is not the last row * / {i = n; / * last row * / j = j +1; / * behind a row * /} else if (j == n && i ! = 1) / * Last but not the first column * / {i = i-1; j = 1; / * first column of the line * /} else if (j! = n && vis [i-1]! [j +1]) / * update the value of the upper right corner seen on tv canada * / {i = i-1; j = j +1; / * right corner * /}! else if (i! = n && j = n && vis [i-1] [j +1]) / * upper right corner has been updated * / i = i +1; / * skip to the next line of the same column * / else if (i == 1 && j == n) / * upper right corner * / i = i +1; / * Skip to the same column of the next line * / ans + +;} for (i = 1; i <= n; i + +) {for (j = 1; j <= n; j + +) printf ("% 4d", a [i] [j]); printf ("\ n");}} return 0;}
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