HDU 482 String

標籤：style   blog   http   color   os   java   io   for   ar   

StringTime Limit: 1000msMemory Limit: 32768KBThis problem will be judged on HDU. Original ID: 4821
64-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. InputThe 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. OutputFor each test case, output the answer in a single line. Sample Input

3 3abcabcbcaabc

Sample Output

2

Source2013 Asia Regional Changchun 解題：字串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 }

View Code

 

HDU 482 String