Sicily 1282 Computer Game

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Brian is an enthusiast of computer games, especially those that simulate virtual reality. Now he is in front of the Star Gate. In order to open the gate he must break the protection as quickly as
he can. Breaking the protection means to match a given code (a sequence of numbers) against the gate protection (a very long sequence of numbers). The starting position of the first occurrence of the code within the sequence opens the gate. Can you help him?

The code is a sequence of at most
60000 integer numbers between 0 and 255. The gate protection contains integer numbers between
0 and 255. Your program must find the first match if there is one, or report the absence of a match.

Input

The text input file contains several data sets. Each data set has the following format:

l    
the length of the code

l    
the sequence of numbers representing the code

l    
the number of integers in the gate protection

l    
the sequence of numbers representing the gate protection 

 code_dimension
integer1 integer2 … integercode_dimension 
protection_dimension
integer1 integer2 … integerprotection_dimension

White spaces may occur freely in the input.

Output

The results must be printed on the standard output. For each given data set, print the result on a separate line. The result is a number that represents the position (starting from zero) of the first
occurrence of the code in the gate protection, or the message no solution
if there is no match.

Sample Input

3
0 1 2
7
2 3 4 0 1 2 5

2
1 4
6
4 1 4 1 4 4

3
1 2 3
7
3 2 1 3 2 255 7

Sample Output

3
1
no solution

Problem Source

ZSUACM Team Member

Solution

给出长度la的序列a，长度不超过60000，给出长度lb的序列b，长度不定，求出a序列在b序列的第一次出现的位置，若没有输出no solution。

字符串匹配，暴力会超时，用KMP就ok啦。

#include <cstdio>
#include <cstring>

int a[1000005], b[100005], la, lb, next[100005];

void kmp()
{
int i, j = 0;

for (i = 1; i <= la; ++i)
{
while (j && a[i] != b[j+1]) j = next[j];
if (a[i] == b[j+1]) ++j;
if (j == lb)
{
printf("%d\n", i - lb);
return;
}
}

printf("no solution\n");
return;
}

int main()
{
while (scanf("%d", &lb) != EOF)
{
int i, t = 0;

for (i = 1; i <= lb; ++i) scanf("%d", &b[i]);
scanf("%d", &la);
for (i = 1; i <= la; ++i) scanf("%d", &a[i]);

  //初始化next序列
next[1] = t;
for (i = 2; i <= lb; ++i)
{
while (t && b[t+1] != b[i]) t = next[t];
if (b[t+1] == b[i]) ++t;
next[i] = t;
}

kmp();
}

return 0;
}