HDU 1711 Number Sequence【kmp】

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

InputThe first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
OutputFor each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input

```2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1```

Sample Output

```6
-1
```

```#include<stdio.h>
int a[1000010],b[10010];
int next[10010];
int n,m;
void getNext()
{
int j,k;
j=0;
k=-1;
next[0]=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
next[++j]=++k;
else k=next[k];
}
}
//返回首次出现的位置
int KMP_Index()
{
int i=0,j=0;
getNext();

while(i<n && j<m)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else j=next[j];

}
if(j==m) return i-m+1;
else return -1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
printf("%d\n",KMP_Index());
}
return 0;
}    ```

kuangbin模板

http://www.cnblogs.com/kuangbin/archive/2012/08/14/2638803.html

hdoj 1711 Number Sequence 【KMP】

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11817    Accepted Submission(s): 5395 Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1],

HDU 1711 Number Sequence (简单KMP)

#include <stdio.h> #include <string.h> int next[10005]; int str1[1000005],str2[10005]; void build_next(int len2) { int i=0,j=-1; next[0] = -1; while (i < len2) { if (j==-1 || str2[i] == str2[j]) { i++; j++; if (str2[i] != str2[j]) { next[i]

HDU 1711 Number Sequence（kmp）

Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1]

HDU 1711 Number Sequence（KMP模板）

http://acm.hdu.edu.cn/showproblem.php?pid=1711 这道题就是一个KMP模板. 1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 5 const int maxn = 1000000+5; 6 7 int n,m; 8 9 int next[maxn]; 10 int a[maxn], b[maxn]; 11 12 void get_next() 13 { 1

hdu 1711 Number Sequence(KMP)

# include <stdio.h> # include <string.h> # include <algorithm> using namespace std; int n,m,next[10010],a[1000010],b[10010]; void Getnext() { int i=0,j=-1; next[0]=-1; while(i<m) { if(j==-1||b[i]==b[j]) i++,j++,next[i]=j; else j=next[

HDU 1711 Number Sequence KMP题解

KMP查找整数数列,不是查找字符串. 原理是一样的,不过把字符串转换为数列,其他基本上是一样的. #include <stdio.h> #include <string.h> const int MAX_N = 1000001; const int MAX_M = 10001; int strN[MAX_N], strM[MAX_M], next[MAX_M], N, M; void getNext() { memset(next, 0, sizeof(int) * M); for