hdu number sentense yt练手1002

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 15   Accepted Submission(s) : 3

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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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

Source

HDU 2007-Spring Programming Contest

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AC代码：

#include <map>#include <queue>#include <cmath>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define maxn 100using namespace std;int N[1000000+10];int M[10000+10];int nex[10000+10];int main(){    int loop,la,lb;    scanf("%d",&loop);    while(loop--){        scanf("%d%d",&la,&lb);        for(int i=0;i<la;++i)            scanf("%d",&N[i]);        for(int i=0;i<lb;++i)            scanf("%d",&M[i]);        int pos=0,j=0;        nex[0]=-1;        nex[1]=0;        for(int i=2;i<lb;++i){            while(pos>=0&&M[pos]!=M[i-1])                pos=nex[pos];            nex[i]=++pos;        }        int p=0,cur=0,t=-1;        while(cur<la){            if(N[cur]==M[p]){                cur++;                p++;            }            else if(p>=0)                p=nex[p];            else {                cur++;                p=0;            }            if(p==lb){                t=cur-lb+1;                break;            }        }        printf("%d\n",t);    }    return 0;}

Compilation Error代码：

#include <iostream>#include<cstdio>#include<string.h>#include<string>using namespace std;int a[1000005];int b[10005];int next[10005];int main(){    int t,la,lb,i;    cin>>t;    while(t--)    {        scanf("%d%d",&la,&lb);        for(i=0; i<la; i++)            scanf("%d",&a[i]);        for(i=0; i<lb; i++)            scanf("%d",&b[i]);        int pos=0;        next[0]=-1;        next[1]=0;        for(i=2; i<lb; ++i)        {            while(pos>=0&&b[pos]!=b[i-1])                pos=next[pos];            next[i]=++pos;        }        int p=0,cur=0,t=-1;        while(cur<la)        {            if(a[cur]==b[p])            {                cur++;                p++;            }            else if(p>=0)                p=next[p];            else            {                cur++;                p=0;            }            if(p==lb)            {                t=cur-lb+1;                break;            }        }        printf("%d\n",t);    }    return 0;}

学习总结：真心不明白哪一点不一样，为什么前个代码对，到了自己打出来一遍却是错的