Simpsons’ Hidden Talents
Problem Description
Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talentin politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.
Input
Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.
Output
Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.
Sample Input
clinton
home
rriemann
marjorie
Sample Output
0
rie 3
Source
HDU 2010-05 Programming Contest
题目类型:KMP
算法分析:本题要求解的是两个字符串中a的前缀和b的后缀相等的最长长度,可以将两个字符串连接在一起,然后求解并输出不超过每个字符串长度的Next[len]中的字符串0~s[Next[len]-1]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 |
#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 7; const int maxn = 100000 + 66; char stra[maxn], strb[maxn]; int Next[maxn]; void Build (char *pat) { int len = strlen (pat); Next[0] = -1; Next[1] = 0; int k = 0; int i; for (i = 2; i <= len; i++) { while (k >= 0 && pat[i-1] != pat[k]) k = Next[k]; Next[i] = ++k; } } int main() { // ifstream cin ("aaa.txt"); while (cin >> stra >> strb) { int lena = strlen (stra), lenb = strlen (strb); strcat (stra, strb); Build (stra); int len = strlen (stra); while (Next[len] > lena || Next[len] > lenb) { len = Next[len]; } if (!Next[len]) cout << "0" << endl; else { len = Next[len]; for (int i = 0; i < len; i++) cout << stra[i]; cout << " " << len << endl; } } return 0; } |
- « 上一篇:hdu2521
- hdu2602:下一篇 »