題目:
Simpsons’ Hidden Talents
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5599 Accepted Submission(s): 2025
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in 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.
The lengths of s1 and s2 will be at most 50000.
描述:給兩個字符串s1,s2,求2的前綴是1的後綴的最大長度
題解:把兩個字符串拼在一起求一遍nxt數組就行了,但要注意長度不能超過兩個字符串的最小長度
代碼:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5+10;
char text[maxn], pattern[maxn];
int nxt[maxn];
int main()
{
//freopen("input.txt", "r", stdin);
while (scanf("%s",text)!=EOF)
{
scanf("%s", pattern);
int len1 = min(strlen(text), strlen(pattern));
strcat(text, pattern);
int len = strlen(text);
for (int i = 0, j = -1; i <= len; i++, j++)
{
nxt[i] = j;
while (~j && text[i] != text[j])
j = nxt[j];
}
for (int i = 0; i < nxt[len] && i < len1; i++)
{
printf("%c", text[i]);
if (i == nxt[len] - 1 || i == len1 - 1)
printf(" ");
}
printf("%d\n", min(nxt[len], len1)); //attention
memset(text, 0, sizeof(text));
memset(pattern, 0, sizeof(pattern));
}
return 0;
}