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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19246 Accepted Submission(s): 8267
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
题解：此题就是如果匹配就输出开始匹配时的数组下标；next数组有两个含义：位置还有长度；
让求串2在串1中首次出现的位置；
代码：
1 #include2 const int MAXN=10010; 3 int a[MAXN*100],b[MAXN],len1,len2,next[MAXN]; 4 void getnext(){ 5 int i=0,j=-1; 6 next[i]=j; 7 while(i
#include#include #include #include #include using namespace std;typedef long long LL;#define mem(x,y) memset(x,y,sizeof(x))#define SI(x) scanf("%d",&x)#define PI(x) printf("%d",x)#define P_ printf(" ")const int INF=0x3f3f3f3f;const int MAXN=1000010;int p[MAXN];int N,M;int s[MAXN];int m[MAXN];void getp(){ int i=0,j=-1; p[0]=-1; while(i
str函数超时：
#include#include #include #include #include using namespace std;typedef long long LL;#define mem(x,y) memset(x,y,sizeof(x))#define SI(x) scanf("%d",&x)#define PI(x) printf("%d",x)#define P_ printf(" ")const int INF=0x3f3f3f3f;const int MAXN=1000010;char s1[MAXN],s2[MAXN];int N,M;/*int s[MAXN];int m[MAXN];void getp(){ int i=0,j=-1; p[0]=-1; while(i