Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22081    Accepted Submission(s): 9443 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   |   We have carefully selected several similar problems for you:   1358  3336  1686  3746  1251  题目大意：给你一个序列a，一个序列b，让你找到b十分存在于a中，存在的话输出b存在于a的起始地址（如果多个情况的话输出最小的），不存在的话输出-1. 解题思路：KMP模板题。 代码如下： #include <cstdio> #include <cstring> int a[1000010]; int b[10010]; int next[10010]; int n,m,ans; void makenext() { int k=0; memset(next,0,sizeof(next)); for(int i=1;i<m;i++) { while(k>0&&b[k]!=b[i]) { k=next[k-1]; } if(b[k]==b[i]) { k++; } next[i]=k; } } void kmp() { int k=0; for(int i=0;i<n;i++) { while(k>0&&b[k]!=a[i]) { k=next[k-1]; } if(b[k]==a[i]) { k++; } if(k==m)//匹配完成 { ans=i-m+2;//举个例子就知道了 1 2 1 2 3 1 2 3 1 3 2 1 2 // 1 2 3 1 3 9-5+2=6 return ; } } } 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]); } makenext(); ans=-1;//初始化ans kmp(); printf("%d\n",ans); } return 0; }