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  • hdu1711 Number Sequence

    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

    很久没有做kmp的题了,这几天要做一下,这是道模板题,直接套模板就行。

    #include<stdio.h>
    #include<string.h>
    int s1[1000005],s2[10006],next[10006],n,m;
    void nextt()
    {
    	int i,j;
    	i=0;j=-1;
    	memset(next,-1,sizeof(next));
    	while(i<m){
    		if(j==-1 || s2[i]==s2[j]){
    			i++;j++;next[i]=j;
    		}
    		else j=next[j];
    	}
    }
    
    int kmp()
    {
    	int i,j;
    	i=0;j=0;
    	while(i<n && j<m){
    		if(j==-1 || s1[i]==s2[j]){
    			i++;j++;
    		}
    		else j=next[j];
    	}
    	if(j>=m){
    		return i+1-m;
    	}
    	else return 0;
    }
    
    int main()
    {
    	int i,j,T;
    	scanf("%d",&T);
    	while(T--)
    	{
    		scanf("%d%d",&n,&m);
    		for(i=0;i<n;i++){
    			scanf("%d",&s1[i]);
    		}
    		for(i=0;i<m;i++){
    			scanf("%d",&s2[i]);
    		}
    		nextt();
    		if(kmp()){
    			printf("%d
    ",kmp());
    		}
    		else printf("-1
    ");
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/herumw/p/9464752.html
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