标签:
链接:http://vjudge.net/problem/51167
分析:
2 2 2
AB
CC
zz
zz
2 8 8
aaaaaaaa
bbbbbbbb
cccccccc
dddddddd
eeeeeeee
ffffffff
gggggggg
hhhhhhhh
abcdefgh
abcdefgh
abcdefgh
abcdefgh
abcdefgh
abcdefgh
abcdefgh
abcdefgh
1 #include <cstdio> 2 3 int n; 4 5 int main() { 6 while (scanf("%d", &n) == 1) { 7 printf("2 %d %d\n", n, n); 8 for (int i = 0; i < n; i++) { 9 for (int j = 0; j < n; j++) 10 putchar(i < 26 ? ‘a‘ + i : ‘A‘ + i - 26); 11 printf("\n"); 12 } 13 14 printf("\n"); 15 for (int i = 0; i < n; i++) { 16 for (int j = 0; j < n; j++) 17 putchar(j < 26 ? ‘a‘ + j : ‘A‘ + j - 26); 18 printf("\n"); 19 } 20 printf("\n"); 21 } 22 return 0; 23 }
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原文地址:http://www.cnblogs.com/XieWeida/p/5862296.html