Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11691 Accepted Submission(s): 5336
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] = b[M].
If there are more than one K exist, output the smallest one.
Input
The 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].
Output
For 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
Source
Recommend
题意：给两个数组a和b，求b在a中出现的的第一个位置，若没有则输出-1，仅仅是数组变成了整型数组，方法与字符数组全然一样。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1000005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int s[maxn],p[maxn],nextval[maxn];
int N,M;
void get_nextval()
{
int i=0,j=-1;
nextval[0]=-1;
while (i<M)
{
if (j==-1||p[i]==p[j])
{
i++;
j++;
if (p[i]!=p[j])
nextval[i]=j;
else
nextval[i]=nextval[j];
}
else
j=nextval[j];
}
}
int KMP()
{
int i=0,j=0;
while (i<N&&j<M)
{
if (j==-1||s[i]==p[j])
{
i++;
j++;
}
else
j=nextval[j];
}
if (j==M)
return i-M+1;
else
return -1;
}
int main()
{
int cas;
scanf("%d",&cas);
while (cas--)
{
scanf("%d%d",&N,&M);
for (int i=0;i<N;i++)
scanf("%d",&s[i]);
for (int i=0;i<M;i++)
scanf("%d",&p[i]);
get_nextval();
int ans=KMP();
printf("%d
",ans);
}
return 0;
}
/*
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
*/