hdu1711 Number Sequence 亦凉 2022-05-30 06:25 139阅读 0赞 # Number Sequence # **Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 33881 Accepted Submission(s): 14126** 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 题目解析: 这是一个简单的模板题目,很水,1A 代码: #include<stdio.h> int a[1000010],b[10010]; int next[10010]; int n,m; void getNext() { int j,k; j=0; k=-1; next[0]=-1; while(j<m) { if(k==-1||b[j]==b[k]) next[++j]=++k; else k=next[k]; } } //返回首次出现的位置 int KMP_Index() { int i=0,j=0; getNext(); while(i<n && j<m) { if(j==-1||a[i]==b[j]) { i++; j++; } else j=next[j]; } if(j==m) return i-m+1; else return -1; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<m;i++) scanf("%d",&b[i]); printf("%d\n",KMP_Index()); } return 0; }
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