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HDU 482 String
String
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 482164-bit integer IO format: %I64d Java class name: Main
Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
(i) It is of length M*L;
(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.
Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".
Your task is to calculate the number of different “recoverable” substrings of S.
(i) It is of length M*L;
(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.
Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".
Your task is to calculate the number of different “recoverable” substrings of S.
Input
The input contains multiple test cases, proceeding to the End of File.
The first line of each test case has two space-separated integers M and L.
The second ine of each test case has a string S, which consists of only lowercase letters.
The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.
The first line of each test case has two space-separated integers M and L.
The second ine of each test case has a string S, which consists of only lowercase letters.
The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.
Output
For each test case, output the answer in a single line.
Sample Input
3 3abcabcbcaabc
Sample Output
2
Source
2013 Asia Regional Changchun
解题：字符串hash+map
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #include <map>14 #define LL long long15 #define ULL unsigned long long16 #define pii pair<int,int>17 #define INF 0x3f3f3f3f18 #define seek 13119 using namespace std;20 const int maxn = 100100;21 map<ULL,int>mp;22 char str[maxn];23 ULL base[maxn],hs[maxn];24 int main() {25 int M,L,len,i,j,ans;26 ULL tmp;27 base[0] = 1;28 for(i = 1; i < maxn; i++) base[i] = base[i-1]*seek;29 while(~scanf("%d%d%s",&M,&L,str)){30 len = strlen(str);31 ans = 0;32 hs[len] = 0;33 for(i = len-1; i >= 0; i--)34 hs[i] = hs[i+1]*seek+str[i]-‘a‘;35 for(i = 0; i < L && i + M*L <= len; i++){36 mp.clear();37 for(j = i; j < i+M*L; j += L){38 tmp = hs[j] - hs[j+L]*base[L];39 mp[tmp]++;40 }41 if(mp.size() == M) ans++;42 for(j = i+M*L; j+L <= len; j += L){43 tmp = hs[j-M*L] - hs[j-M*L+L]*base[L];44 mp[tmp]--;45 if(!mp[tmp]) mp.erase(tmp);46 tmp = hs[j] - hs[j+L]*base[L];47 mp[tmp]++;48 if(mp.size() == M) ans++;49 }50 }51 printf("%d\n",ans);52 }53 return 0;54 }
HDU 482 String
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