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Crazy Search
Time Limit: 1000MS | Memory Limit: 65536K |
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.
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.
题意
给一个字符串,问在该串中总共出现了几种长度为n的子串
哈希。直接搞,最好是双哈希,单哈希不一定能过。

1 #include <iostream> 2 #include <sstream> 3 #include <ios> 4 #include <iomanip> 5 #include <functional> 6 #include <algorithm> 7 #include <vector> 8 #include <string> 9 #include <list> 10 #include <queue> 11 #include <deque> 12 #include <stack> 13 #include <set> 14 #include <map> 15 #include <cstdio> 16 #include <cstdlib> 17 #include <cmath> 18 #include <cstring> 19 #include <climits> 20 #include <cctype> 21 using namespace std; 22 #define XINF INT_MAX 23 #define INF 0x3FFFFFFF 24 #define MP(X,Y) make_pair(X,Y) 25 #define PB(X) push_back(X) 26 #define REP(X,N) for(int X=0;X<N;X++) 27 #define REP2(X,L,R) for(int X=L;X<=R;X++) 28 #define DEP(X,R,L) for(int X=R;X>=L;X--) 29 #define CLR(A,X) memset(A,X,sizeof(A)) 30 #define IT iterator 31 typedef long long ll; 32 typedef pair<int,int> PII; 33 typedef vector<PII> VII; 34 typedef vector<int> VI; 35 const int B=16000057; 36 char s[B]; 37 bool a[B]; 38 bool c[26]; 39 map<char,int>m; 40 int main() 41 { 42 ios::sync_with_stdio(false); 43 int n,nc; 44 while(scanf("%d%d",&n,&nc)!=EOF){ 45 scanf("%s",s); 46 int len=strlen(s); 47 int b=0; 48 memset(a,0,sizeof(a)); 49 for(int i=0;i<len;i++) 50 { 51 if(c[s[i]-'a']==0)c[s[i]-'a']=1; 52 } 53 int t=0; 54 for(int i=0;i<26;i++) 55 { 56 if(c[i]) 57 { 58 m[i+'a']=t; 59 t++; 60 } 61 } 62 b=1; 63 int bl=0; 64 //for(int i=0;i<len;i++)cout<<m[s[i]]<<endl; 65 for(int i=0;i<n;i++)b=(b*nc)%B; 66 for(int i=0;i<n;i++) 67 { 68 bl=((bl*nc)%B+m[s[i]])%B; 69 } 70 a[bl]=1; 71 for(int i=n;i<len;i++) 72 { 73 bl=((bl*nc-m[s[i-n]]*b+B)%B+m[s[i]])%B; 74 a[bl]=1; 75 } 76 int ans=0; 77 for(int i=0;i<B;i++) 78 { 79 if(a[i])ans++; 80 } 81 printf("%d ",ans); 82 } 83 return 0; 84 }