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题目:
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
大意:
有a,b两个整数数组,在a数组里面找到第一个匹配b数组的位置。
思路:
KMP模板题
代码:
/*by kzl*/#include#include #include #include #include #include #include #include using namespace std;const int maxx = 1e6+500;const int INF = 0x3f3f3f3f;typedef long long LL;int n,m,t;int a[maxx],b[maxx];int fail[maxx];void getfail(int len){ int i=0,j=-1; fail[0] = -1; while(i!=len){ if(j==-1||b[i]==b[j]){ fail[++i] = ++j; } else{ j = fail[j]; } }}int getindex(int len,int len1){ int i=0,j=0; while(i!=len){ if(j==-1||a[i]==b[j]){ i++;j++; if(j==len1)return i-len1+1; } else { j = fail[j]; } } return -1;}int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); for(int i=0;i
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