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HDU——T 1711 Number Sequence
http://acm.hdu.edu.cn/showproblem.php?pid=1711
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29129 Accepted Submission(s): 12254
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
213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1
Sample Output
6-1
模板练习题
1 #include <algorithm> 2 #include <cstdio> 3 4 using namespace std; 5 6 const int N(1000000+5); 7 const int M(10000+626); 8 int l1,l2,a[N],b[N],p[N]; 9 10 inline void Get_next()11 {12 for(int j=0,i=2;i<=l2;i++)13 {14 if(j>0&&b[i]!=b[j+1]) j=p[j];15 if(b[i]==b[j+1]) j++;16 p[i]=j;17 }18 }19 inline void kmp()20 {21 for(int j=0,i=1;i<=l1;i++)22 {23 if(j>0&&a[i]!=b[j+1]) j=p[j];24 if(a[i]==b[j+1]) j++;25 if(j==l2)26 {27 printf("%d\n",i-j+1);28 return ;29 }30 }31 puts("-1");32 }33 34 int main()35 {36 int t;scanf("%d",&t);37 for(;t--;)38 {39 scanf("%d%d",&l1,&l2);40 for(int i=1;i<=l1;i++) scanf("%d",a+i);41 for(int i=1;i<=l2;i++) scanf("%d",b+i);42 Get_next();43 kmp();44 }45 return 0;46 }
HDU——T 1711 Number Sequence
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