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
Author
ZHOU, Yuchen
解题：在SAM上dp
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 351000;
4 struct SAM {
5 struct node {
6 int son[26],f,len;
7 void init() {
8 f = -1;
9 len = 0;
10 memset(son,-1,sizeof son);
11 }
12 } s[maxn<<1];
13 int tot,last;
14 void init() {
15 tot = last = 0;
16 s[tot++].init();
17 }
18 int newnode() {
19 s[tot].init();
20 return tot++;
21 }
22 void extend(int c){
23 int np = newnode(),p = last;
24 s[np].len = s[p].len + 1;
25 while(p != -1 && s[p].son[c] == -1){
26 s[p].son[c] = np;
27 p = s[p].f;
28 }
29 if(p == -1) s[np].f = 0;
30 else{
31 int q = s[p].son[c];
32 if(s[p].len + 1 == s[q].len) s[np].f = q;
33 else{
34 int nq = newnode();
35 s[nq] = s[q];
36 s[nq].len = s[p].len + 1;
37 s[q].f = s[np].f = nq;
38 while(p != -1 && s[p].son[c] == q){
39 s[p].son[c] = nq;
40 p = s[p].f;
41 }
42 }
43 }
44 last = np;
45 }
46 } sam;
47 queue<int>q;
48 int du[maxn],dp[maxn][32],ans[maxn],k;
49 char str[maxn];
50 int main(){
51 while(~scanf("%s%d",str,&k)) {
52 memset(du,0,sizeof du);
53 memset(ans,0,sizeof ans);
54 sam.init();
55 for(int i = 0; str[i]; ++i) sam.extend(str[i] - ‘0‘);
56 for(int i = 0; i < sam.tot; ++i) {
57 memset(dp[i],0,sizeof dp[i]);
58 for(int j = 0; j < 10; ++j) if(sam.s[i].son[j] != -1) ++du[sam.s[i].son[j]];
59 }
60 //cout<<du[0]<<endl;
61 while(!q.empty());
62 for(int i = 0; i < sam.tot; ++i) if(!du[i]) q.push(i);
63 dp[0][0] = 1;
64 while(!q.empty()) {
65 int u = q.front();
66 q.pop();
67 for(int i = 0; i < 10; ++i) {
68 int v = sam.s[u].son[i];
69 if(v == -1) continue;
70 if(--du[v] == 0) q.push(v);
71 if(u == 0 && i == 0) continue;
72 for(int j = 0; j < k; ++j)
73 dp[v][(j*10 + i)%k] += dp[u][j];
74 }
75 for(int i = 0; i < k; ++i)
76 ans[i] += dp[u][i];
77 }
78 ans[0]--;
79 for(int i = 0; str[i]; ++i)
80 if(str[i] == ‘0‘) {
81 ans[0]++;
82 break;
83 }
84 for(int i = 0; i < k-1; ++i)
85 printf("%d ",ans[i]);
86 printf("%d\n",ans[k-1]);
87 }
88 return 0;
89 }