UVA 12604 - Caesar Cipher
题意：给一个字母表s，一个标准串w，一个密文s，问w是否可能在密文的原文中出现且仅出现一次
#include using namespace std; const int maxn = 6e5; char a[maxn], b[maxn], o[maxn]; int c[300]; int pi[maxn], f[maxn]; void getnext(int n) { pi[1] = 0; for (int i = 2, j = 0; i <= n; i++) { while (j > 0 && a[i] != a[j + 1]) j = pi[j]; if (a[i] == a[j + 1]) j++; pi[i] = j; } } bool solve(int n, int m) { int flag = 0; for (int i = 1, j = 0; i <= m; i++) { while (j > 0 && (j == n || b[i] != a[j + 1])) j = pi[j]; if (b[i] == a[j + 1]) j++; f[i] = j; if (f[i] == n) { flag++; } } return flag == 1; } vector ans; int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); int _; scanf("%d", &_); char s[100]; while (_--) { ans.clear(); memset(c, 0, sizeof(c)); scanf("%s %s %s", s, o + 1, b + 1); int na = strlen(o + 1), nt = strlen(b + 1); int ns = strlen(s); for (int i = 0; i < ns; i++) { c[s[i]] = i; } int cnt = 0; for (int k = 0; k < ns; k++) { for (int i = 1; i <= na; i++) { a[i] = s[(c[o[i]] + k) % ns]; } a[na + 1] = '\0'; getnext(na); if (solve(na, nt)) ans.push_back(k), cnt++; } if (cnt == 0) printf("no solution\n"); else if (cnt == 1) printf("unique: %d\n", ans[0]); else { printf("ambiguous:"); for (auto v:ans) { printf(" %d", v); } printf("\n"); } } return 0; }