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 = 351000;
struct SAM {
    struct node {
        int son[26],f,len;
        void init() {
            f = -1;
            len = 0;
            memset(son,-1,sizeof son);
        }
    } s[maxn<<1];
    int tot,last;
    void init() {
        tot = last = 0;
        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 + 1;
        while(p != -1 && s[p].son[c] == -1){
            s[p].son[c] = np;
            p = s[p].f;
        }
        if(p == -1) s[np].f = 0;
        else{
            int q = s[p].son[c];
            if(s[p].len + 1 == s[q].len) s[np].f = q;
            else{
                int nq = newnode();
                s[nq] = s[q];
                s[nq].len = s[p].len + 1;
                s[q].f = s[np].f = nq;
                while(p != -1 && 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][32],ans[maxn],k;
char str[maxn];
int main(){
    while(~scanf("%s%d",str,&k)) {
        memset(du,0,sizeof du);
        memset(ans,0,sizeof ans);
        sam.init();
        for(int i = 0; str[i]; ++i) sam.extend(str[i] - '0');
        for(int i = 0; i < sam.tot; ++i) {
            memset(dp[i],0,sizeof dp[i]);
            for(int j = 0; j < 10; ++j) if(sam.s[i].son[j] != -1) ++du[sam.s[i].son[j]];
        }
        //cout<<du[0]<<endl;
        while(!q.empty());
        for(int i = 0; i < sam.tot; ++i) if(!du[i]) q.push(i);
        dp[0][0] = 1;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            for(int i = 0; i < 10; ++i) {
                int v = sam.s[u].son[i];
                if(v == -1) continue;
                if(--du[v] == 0) q.push(v);
                if(u == 0 && i == 0) continue;
                for(int j = 0; j < k; ++j)
                    dp[v][(j*10 + i)%k] += dp[u][j];
            }
            for(int i = 0; i < k; ++i)
                ans[i] += dp[u][i];
        }
        ans[0]--;
        for(int i = 0; str[i]; ++i)
            if(str[i] == '0') {
                ans[0]++;
                break;
            }
        for(int i = 0; i < k-1; ++i)
            printf("%d ",ans[i]);
        printf("%d
",ans[k-1]);
    }
    return 0;
}