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Number Sequence Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19556 Accepted Submission(s): 8389 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 Source HDU 2007-Spring Programming Contest Recommendlcy
可以用单个数字输入抵消空格的影响 不直接用string 进行kmp算法
#include#include #include using namespace std;int nexta[4000100];int a[1000100],b[1000010];//a是子串b是模板 void getnext(int a[],long long h){ int i=0,j=-1; nexta[0]=-1; while(i >n; while(n--) { cin>>k>>h; for(i=0;i >b[i]; for(i=0;i >a[i]; getnext(a,h); cout< <
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