POJ 1200 Crazy Search

Crazy Search Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10954   Accepted: 2936 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 4 daababac Sample Output 5 Hint Huge input,scanf is recommended. Source Southwestern Europe 2002

 

/* http://acm.pku.edu.cn/JudgeOnline/problem?id=1200 哈希，cn进制表示 一开始在循环里每次都求strlen(input)，结果超时， 预先求好strlen就OK了，输入规模太大了 */ #include <iostream> #define MAX_N 16000000 #define MAX_L 200000000 using namespace std; char input[MAX_L + 1]; int hash[MAX_N + 1]; int occur[130], index[130], indexNum; int main() { //cout<<int('Z')<<" "<<int('z')<<endl; int i, j, hashVal, countv = 0, n, nc; scanf("%d%d%s", &n, &nc, input); //统计字符的位置 indexNum = 0; for(i = 0; i < strlen(input); i++) { if(occur[input[i]] == 0) indexNum++; occur[input[i]]++; if(indexNum == nc) break; } int p = 0; for(i = 0; i < 130; i++) { if(occur[i] != 0) index[i] = p++; } countv = 0; int len = strlen(input); for(i = 0; i <= len - n; i++) { hashVal = 0; for(j = i; j < i + n; j++) hashVal = hashVal * nc + index[input[j]]; if(hash[hashVal] == 0) countv++; hash[hashVal]++; } printf("%d/n", countv); return 0; }