Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22070    Accepted Submission(s): 9437 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 注意输出的时候要在待输出的位置加1，因为题目是默认是从1开始的，我们是从0开始输入的；

KMP 算法：

#include<stdio.h> #include<string.h> int a[1000010]; int b[1000010]; int next[1000010]; void makenext(int m) { int j=0; next[0]=0; for(int i=1;i<m;i++) { while(j&&b[j]!=b[i]) { j=next[j-1]; } if(b[j]==b[i]) { j++; } next[i]=j; } } void KMP(int n,int m) { int j=0; for(int i=0;i<n;i++) { while(j&&a[i]!=b[j]) { j=next[j-1]; } if(a[i]==b[j]) { j++; } if(j>=m) { printf("%d\n",i-m+1+1); return; } } printf("-1\n"); } int main() { int t,n,m; 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]); } makenext(m); KMP(n,m); } return 0; }

MP算法： #include<stdio.h> #include<string.h> int n,m; int a[1000010]; int b[10010]; int next[10010];//f[i]表示长度为i的子串前缀和后缀公共部分的最大长度；即最大公共长度；从1开始存的； void get() { int i,j; next[0]=next[1]=0; for(int i=1;i<m;i++) { j=next[i];//j每次都表示next[i]的值即最大公共长度，同时也表示需要比较的下一个位置； while( j && b[j]!=b[i]) j=next[j];//分割长度为j的字符串； next[i+1]=b[j]==b[i]?j+1:0; } } void find() { int i,j=0; for(int i=0;i<n;i++) { while( j && a[i]!=b[j])//如果j==0则j不再动，让i移动，知道找到一个i使a[i]==b[0] { //待比较的元素不匹配，移动j直到匹配； j=next[j]; } if(a[i]==b[j])//相等则j往后移； j++; if(j>=m)//匹配成功； { printf("%d\n",i-m+1+1); return; } } printf("-1\n");//循环结束后匹配不成功，输出-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]); } get(); find(); } return 0; }