原题链接：https://vjudge.net/problem/UVA-129 分类：回溯法 备注：隐性回溯,字符串
代码如下：
#include<cstdio> using namespace std; const char a[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"; int n, L, cnt; char ans[90]; bool over; bool check(int pos) { int len = 1, tot = pos + 1; while (2 * len <= tot) { int last = pos - len; bool flag = true; for (int i = pos; i > last; i--) { if (ans[i] != ans[i - len]) { flag = false; break; } } if (flag)return false; len++; } return true; } void solve(int pos) {//在下标为pos处添加字符 if (over)return; if (cnt == n) { for (int i = 0, num = 0; i < pos; i++) { printf("%c", ans[i]); if ((i + 1) % 4 == 0 && i + 1 != pos) { num++; if (num % 16 == 0) printf("\n"); else printf(" "); } } printf("\n%d\n", pos); over = true; return; } for (int i = 0; i < L; i++) { ans[pos] = a[i]; if (check(pos)) { cnt++; solve(pos + 1); if (over)return; } } } int main(void) { while (scanf("%d %d", &n, &L) == 2 && L) { over = false; cnt = 0; solve(0); } return 0; }