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题意:给它定一个n,让你输出一个n*n的矩阵,使得整个矩阵,每行,每列,对角线和都是奇数。 析:这个题可以用n阶奇幻方来解决,当然也可以不用,如果不懂,请看:http://www.cnblogs.com/dwtfukgv/articles/5797527.html 剩下的就很简单了。 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 100000000000000000; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 3e5 + 5; const int mod = 1e9 + 7; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; int n, m; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } inline LL Max(LL a, LL b){ return a < b ? b : a; } inline LL Min(LL a, LL b){ return a > b ? b : a; } inline int Max(int a, int b){ return a < b ? b : a; } inline int Min(int a, int b){ return a > b ? b : a; } int a[55][55]; int main(){ while(scanf("%d", &n) == 1){ int j = n/2, num = 1, i = 0; while(num != n*n + 1){ int ii = (i % n + n) % n; int jj = (j % n + n) % n; a[ii][jj] = num; if(num % n == 0) ++i; else --i, ++j; ++num; } for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) if(j == n-1) printf("%d\n", a[i][j]); else printf("%d ", a[i][j]); } return 0; }
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