• hdu 1711 Number Sequence


    hdu 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)

    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
     题目大意:输出首次匹配位置
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int T,n,m,ans;
    int a[1000001],b[10001],f[10001];
    void getnext()
    {
        memset(f,0,sizeof(f));
        for(int i=1;i<m;i++)
        {
            int j=f[i];
            while(j&&b[i]!=b[j]) j=f[j];
            f[i+1]= b[i]==b[j] ? j+1 : 0;
        }
    }
    void getans()
    {
        ans=0;int j=0;
        for(int i=0;i<n;i++)
        {
            while(j&&a[i]!=b[j]) j=f[j];
            if(a[i]==b[j]) j++;
            if(j==m) {printf("%d
    ",i-j+2);    return;    }
        }
        printf("-1
    ");
    }
    int main()
    {
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m);
            for(int i=0;i<n;i++) scanf("%d",&a[i]);
            for(int j=0;j<m;j++) scanf("%d",&b[j]);
            getnext();
            getans();
        }
    } 
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/6483192.html
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