str2int
Time Limit: 3000ms
Memory Limit: 131072KB
This problem will be judged on HDU. Original ID: 4436
64-bit integer IO format: %I64d Java class name: Main
In this problem, you are given several strings that contain only digits from ‘0‘ to ‘9‘, inclusive.
An example is shown below.
101
123
The set S of strings is consists of the N strings given in the input file, and all the possible substrings of each one of them.
It‘s boring to manipulate strings, so you decide to convert strings in S into integers.
You can convert a string that contains only digits into a decimal integer, for example, you can convert "101" into 101, "01" into 1, et al.
If an integer occurs multiple times, you only keep one of them.
For example, in the example shown above, all the integers are 1, 10, 101, 2, 3, 12, 23, 123.
Your task is to calculate the remainder of the sum of all the integers you get divided by 2012.
Input
There are no more than 20 test cases.
The test case starts by a line contains an positive integer N.
Next N lines each contains a string consists of one or more digits.
It‘s guaranteed that 1≤N≤10000 and the sum of the length of all the strings ≤100000.
The input is terminated by EOF.
Output
An integer between 0 and 2011, inclusive, for each test case.
Sample Input
5 101 123 09 000 1234567890
Sample Output
202
Source
2012 Asia Tianjin Regional Contest
解题：后缀自动机大法好
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int mod = 2012;
4 const int maxn = 210001;
5 struct node {
6 int son[11],f,len;
7 void init() {
8 f = -1;
9 len = 0;
10 memset(son,-1,sizeof son);
11 }
12 };
13 struct SAM {
14 node e[maxn<<1];
15 int tot,last;
16 int newnode(int len = 0) {
17 e[tot].init();
18 e[tot].len = len;
19 return tot++;
20 }
21 void init() {
22 tot = last = 0;
23 newnode(0);
24 }
25 void extend(int c) {
26 int p = last,np = newnode(e[p].len + 1);
27 while(p != -1 && e[p].son[c] == -1) {
28 e[p].son[c] = np;
29 p = e[p].f;
30 }
31 if(p == -1) e[np].f = 0;
32 else{
33 int q = e[p].son[c];
34 if(e[p].len + 1 == e[q].len) e[np].f = q;
35 else{
36 int nq = newnode();
37 e[nq] = e[q];
38 e[nq].len = e[p].len + 1;
39 e[np].f = e[q].f = nq;
40 while(p != -1 && e[p].son[c] == q){
41 e[p].son[c] = nq;
42 p = e[p].f;
43 }
44 }
45 }
46 last = np;
47 }
48 }sam;
49 char str[maxn];
50 int cnt[maxn<<1],c[maxn<<1],sum[maxn<<1],sa[maxn<<1];
51 int main() {
52 int n,len;
53 while(~scanf("%d",&n)){
54 sam.init();
55 for(int i = len = 0; i < n; ++i){
56 scanf("%s",str + len);
57 len = strlen(str);
58 if(i + 1 < n) str[len++] = ‘0‘ + 10;
59 }
60 for(int i = 0; i < len; ++i)
61 sam.extend(str[i] - ‘0‘);
62 memset(c,0,sizeof c);
63 memset(cnt,0,sizeof cnt);
64 memset(sum,0,sizeof sum);
65 for(int i = 0; i < sam.tot; ++i) c[sam.e[i].len]++;
66 for(int i = 1; i <= len; ++i) c[i] += c[i-1];
67 for(int i = sam.tot-1; i >= 0; --i) sa[--c[sam.e[i].len]] = i;
68 int ret = 0;
69 cnt[0] = 1;
70 for(int i = 0; i < sam.tot; ++i){
71 int x = sa[i];
72 for(int j = 0; j < 10; ++j){
73 if(x == 0 && j == 0) continue;
74 cnt[sam.e[x].son[j]] += cnt[x];
75 sum[sam.e[x].son[j]] += sum[x]*10 + cnt[x]*j;
76 cnt[sam.e[x].son[j]] %= mod;
77 sum[sam.e[x].son[j]] %= mod;
78 }
79 ret = (ret + sum[x])%mod;
80 }
81 printf("%d\n",ret);
82 }
83 return 0;
84 }