解题思路:
打个表,先把图形放在数组里,然后输出输出输出...
注意事项:
参考代码:
#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <algorithm> #include <string> #include <vector> #include <set> #include <map> #include <math.h> #define N 41 using namespace std; char G[N][N]; void clearSingleGraph() { for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) G[i][j] = ' '; } void getSingleGraph(int n) { clearSingleGraph(); for (int i = 1; i <= n; i++) G[i][i] = G[i][n - i + 1] = 'X'; return; } //参数1:重复度 //参数2:是否为之后的图形 void showSingleGraph(int n, bool flag) { int i = 0, j = 0; if (flag) i = 2; else i = j = 1; for (; i <= n; i++) { for (j = 1; j <= n; j++) cout << G[i][j]; cout << endl; } } //参数1:单个图形的规模 //参数2:重复度 void showGraph(int m, int n) { getSingleGraph(m); for (int i = 1; i <= n; i++) { if (i == 1) showSingleGraph(m, false); else showSingleGraph(m, true); } cout << endl; } int main(int argc, char** argv) { //showGraph(5, 4); int times = 0; cin >> times; while (times--) { int m = 0, n = 0; cin >> m >> n; showGraph(m, n); } return 0; } 【思路二】 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 45; int n,a[maxn],b[maxn]; void print_DNA(int a, int b){ for(int k = 1; k <= b; k++){ for(int i = 1; i < a; i++){ for(int j = 1; j <= a; j++){ if(i==j || i+j==a+1) printf("X"); else printf(" "); } printf("\n"); } } for(int j = 1; j <= a; j++){ if(j==1 || j==a) printf("X"); else printf(" "); } } int main (void){ scanf("%d", &n); for(int i = 0; i < n; i++){ scanf("%d%d", &a[i], &b[i]); } for(int i = 0; i < n; i++){ print_DNA(a[i],b[i]); printf("\n\n"); } return 0; }
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