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hdu——1711Number Sequence(kmp专练)
阅读量:4047 次
发布时间:2019-05-25

本文共 1526 字,大约阅读时间需要 5 分钟。

Number Sequence
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19556    Accepted Submission(s): 8389
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
 HDU 2007-Spring Programming Contest  
Recommend

lcy

可以用单个数字输入抵消空格的影响 不直接用string 进行kmp算法

#include
#include
#include
using namespace std;int nexta[4000100];int a[1000100],b[1000010];//a是子串b是模板 void getnext(int a[],long long h){ int i=0,j=-1; nexta[0]=-1; while(i
>n; while(n--) { cin>>k>>h; for(i=0;i
>b[i]; for(i=0;i
>a[i]; getnext(a,h); cout<
<

转载地址:http://dtfci.baihongyu.com/

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