• hdu-5918 Sequence I(kmp)


    题目链接:

    Sequence I

    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 687    Accepted Submission(s): 262


    Problem Description
    Mr. Frog has two sequences a1,a2,,an and b1,b2,,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,,bm is exactly the sequence aq,aq+p,aq+2p,,aq+(m1)p where q+(m1)pn and q1.
     
    Input
    The first line contains only one integer T100, which indicates the number of test cases.

    Each test case contains three lines.

    The first line contains three space-separated integers 1n106,1m106 and 1p106.

    The second line contains n integers a1,a2,,an(1ai109).

    the third line contains m integers b1,b2,,bm(1bi109).
     
    Output
    For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.
     
    Sample Input
    2 6 3 1 1 2 3 1 2 3 1 2 3 6 3 2 1 3 2 2 3 1 1 2 3
     
    Sample Output
    Case #1: 2
    Case #2: 1
     
    题意:
     
    问a数组里面匹配的b有多少个;
     
    思路:
     
    把a里面间隔p的拿出来,然后用b数组跑kmp求;
     
    AC代码:
     
    #include <bits/stdc++.h>
    using namespace std;
    const int maxn=1e6+10;
    int n,m,p,a[maxn],b[maxn],c[maxn],nex[maxn];
    inline void makenext()
    {
        int k=-1,j=0;nex[0]=-1;
        while(j<m)
        {
            if(k==-1||b[j]==b[k])
            {
                j++;k++;
                nex[j]=k;
            }
            else k=nex[k];
        }
    }
    inline int kmp(int l)
    {
        int pb=0,pc=0,ans=0;
        while(pb<m&&pc<l)
        {
            if(pb==-1||b[pb]==c[pc])pb++,pc++;
            else pb=nex[pb];
            if(pb==m)ans++,pb=nex[pb];
        }
        return ans;
    }
    int main()
    {
        int t,Case=0;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d%d",&n,&m,&p);
            for(int i=0;i<n;i++)scanf("%d",&a[i]);
            for(int i=0;i<m;i++)scanf("%d",&b[i]);
            makenext();
            int ans=0;
            for(int i=0;i<p;i++)
            {
                int cnt=0;
                for(int j=i;j<n;j+=p)c[cnt++]=a[j];
                ans+=kmp(cnt);
            }
            printf("Case #%d: %d
    ",++Case,ans);
        }    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5939429.html
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