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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 第一道kmp 是一道模板題 不過還是wa了兩發 1.不能開next數組 和系統自帶名字重名 2.一開始我是把一個個數據轉化成一條字符串來進行kmp 但是顯然是錯的QWQ 改成數組形式就行了

#include<bits/stdc++.h>
using namespace std;
//input
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);i--)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m);
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define inf 0x3f3f3f3f
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
int nex[100000+5];
int lenp,lens;
int p[10000+5];//可以是char 也可以是string
int s[1000000+5];
void getnext()
{
    nex[0]=-1;
    int k=-1,j=0;
    while(j<lenp-1)
    {
        if(k==-1||p[j]==p[k])
            nex[++j]=++k;
        else k=nex[k];
    }
}
int kmp()
{   
    //lens=
    //lenp=
    int j=0;
    int i=0;
    while(i<lens&&j<lenp)
    {
        if(s[i]==p[j]||j==-1)
        {
            i++;
            j++;
        }
        else
            j=nex[j];

        if(j==lenp)
        {
            return i-j+1;
        }
    }
    return -1;
}
int main()
{
    int cas;
    RI(cas);
    while(cas--)
    {
        RII(lens,lenp);
        rep(i,0,lens-1)
           RI(s[i]);
        rep(i,0,lenp-1)
            RI(p[i]);

        getnext();
        cout<<kmp()<<endl;
    }
    return 0;
}

View Code

Number Sequence kmp