Crazy Search
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 27536 | Accepted: 7692 |
Description
Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in the text. As you soon will discover, you really need the help of a computer and a good algorithm to solve such a puzzle. Your task is to write a program that given the size, N, of the substring, the number of different characters that may occur in the text, NC, and the text itself, determines the number of different substrings of size N that appear in the text. As an example, consider N=3, NC=4 and the text "daababac". The different substrings of size 3 that can be found in this text are: "daa"; "aab"; "aba"; "bab"; "bac". Therefore, the answer should be 5.
Input
The first line of input consists of two numbers, N and NC, separated by exactly one space. This is followed by the text where the search takes place. You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions.
Output
The program should output just an integer corresponding to the number of different substrings of size N found in the given text.
Sample Input
3 4daababac
Sample Output
5
Hint
Huge input,scanf is recommended.
思路：把长度为n的子串hash成nc进制数，避免了冲突。
1 //2016.9.4 2 #include3 #include 4 #include 5 #define N 16000005 6 7 using namespace std; 8 9 char str[N];10 bool Hash[N];11 int id[500];12 13 int main()14 {15 int n, nc, ans, cnt;16 while(scanf("%d%d", &n, &nc)!=EOF)17 {18 cnt = ans = 0;19 memset(Hash, false, sizeof(Hash));20 memset(id, -1, sizeof(id));21 scanf("%s", str);22 int len = strlen(str);23 for(int i = 0; i < len && cnt < nc; i++)//对str出现的字符进行编号，使之转换成数字24 {25 if(id[str[i]] != -1)continue;26 id[str[i]] = cnt++;27 }28 for(int i = 0; i < len-n+1; i++)//把长度为n的子串hash成nc进制数29 {30 int tmp = 0;31 for(int j = i; j < i+n; j++)32 tmp = tmp*nc+id[str[j]];33 if(Hash[tmp])continue;34 ans++;35 Hash[tmp] = true;36 }37 printf("%d\n", ans);38 }39 40 return 0;41 }