ZOJ 3891 K-hash

K-hash

Time Limit: 2000ms

Memory Limit: 131072KB

This problem will be judged on ZJU. Original ID: 3891
64-bit integer IO format: %lld      Java class name: Main

 

K-hash is a simple string hash function. It encodes a string Sconsist of digit characters into a K-dimensional vector (h0, h1, h2,... , hK-1). If a nonnegative number x occurs in S, then we call x is S-holded. And hi is the number of nonnegative numbers which are S-holded and congruent with i modulo K, for i from 0 to K-1.

For example, S is "22014" and K=3. There are 12 nonnegative numbers are "22014"-holded: 0, 1, 2, 4, 14, 20, 22, 201, 220, 2014, 2201 and 22014. And three of them, 0, 201 and 22014, are congruent with 0 modulo K, so h0=3. Similarly, h1=5 (1, 4, 22, 220 and 2014 are congruent with 1 modulo 3), h2=4(2, 14, 20 and 2201 are congruent with 2 modulo 3). So the 3-hash of "22014" is (3, 5, 4).

Please calculate the K-hash value of the given string S.

Input

There are multiple cases. Each case is a string S and a integer number K. (S is a string consist of '0', '1', '2', ... , '9' , 0< |S| ≤ 50000, 0< K≤ 32)

Output

For each case, print K numbers (h0, h1, h2,... , hK-1 ) in one line.

Sample Input

123456789 10
10203040506007 13
12345678987654321 2
112123123412345123456123456712345678123456789 17
3333333333333333333333333333 11

Sample Output

0 1 2 3 4 5 6 7 8 9
3 5 5 4 3 2 8 3 5 4 2 8 4
68 77
57 58 59 53 49 57 60 55 51 45 59 55 53 49 56 42 57
14 0 0 14 0 0 0 0 0 0 0

Source

ZOJ Monthly, July 2023

Author

ZHOU, Yuchen

 

解题：在SAM上dp

 

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 struct SAM {
     struct node {
         int son[],f,len;
         void init() {
             f = -;
             len = ;
             memset(son,-,sizeof son);
         }
     } s[maxn<<];
     int tot,last;
     void init() {
         tot = last = ;
         s[tot++].init();
     }
     int newnode() {
         s[tot].init();
         return tot++;
     }
     void extend(int c){
         int np = newnode(),p = last;
         s[np].len = s[p].len + ;
         while(p != - && s[p].son[c] == -){
             s[p].son[c] = np;
             p = s[p].f;
         }
         if(p == -) s[np].f = ;
         else{
             int q = s[p].son[c];
             if(s[p].len +  == s[q].len) s[np].f = q;
             else{
                 int nq = newnode();
                 s[nq] = s[q];
                 s[nq].len = s[p].len + ;
                 s[q].f = s[np].f = nq;
                 while(p != - && s[p].son[c] == q){
                     s[p].son[c] = nq;
                     p = s[p].f;
                 }
             }
         }
         last = np;
     }
 } sam;
 queue<int>q;
 int du[maxn],dp[maxn][],ans[maxn],k;
 char str[maxn];
 int main(){
     while(~scanf("%s%d",str,&k)) {
         memset(du,,sizeof du);
         memset(ans,,sizeof ans);
         sam.init();
         for(int i = ; str[i]; ++i) sam.extend(str[i] - '');
         for(int i = ; i < sam.tot; ++i) {
             memset(dp[i],,sizeof dp[i]);
             for(int j = ; j < ; ++j) if(sam.s[i].son[j] != -) ++du[sam.s[i].son[j]];
         }
         //cout<<du[0]<<endl;
         while(!q.empty());
         for(int i = ; i < sam.tot; ++i) if(!du[i]) q.push(i);
         dp[][] = ;
         while(!q.empty()) {
             int u = q.front();
             q.pop();
             for(int i = ; i < ; ++i) {
                 int v = sam.s[u].son[i];
                 if(v == -) continue;
                 if(--du[v] == ) q.push(v);
                 if(u ==  && i == ) continue;
                 for(int j = ; j < k; ++j)
                     dp[v][(j* + i)%k] += dp[u][j];
             }
             for(int i = ; i < k; ++i)
                 ans[i] += dp[u][i];
         }
         ans[]--;
         for(int i = ; str[i]; ++i)
             if(str[i] == '') {
                 ans[]++;
                 break;
             }
         for(int i = ; i < k-; ++i)
             printf("%d ",ans[i]);
         printf("%d\n",ans[k-]);
     }
     return ;
 }