hdu 4436 str2int (SAM)（待补）

参考：http://blog.csdn.net/u013654696/article/details/40661797

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

给n个只包含数字的字符串， 问这n个字符串能分解为多少种不同的数字， 求出这些数字的和mod2012

先把n个串用10连接起来， 然后构造sam。 然后计数的时候可以从拓扑序从小到大计数，（以前一直以为只能从大到小。。。）， cnt[i]表示这个节点上有多少种数， sum[i]表示这个节点之前的能构成的数的和是多少，转移就是直接转移到儿子上就可以了。。。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define mnx 200020
#define mod 2012

int ch[mnx << 1][12], par[mnx << 1], val[mnx << 1], sz;
int cnt[mnx << 1], sum[mnx << 1];
int root, lst;

int creat(int v) {
++sz;
val[sz] = v;
par[sz] = 0;
cnt[sz] = 0;
sum[sz] = 0;
memset(ch[sz], 0, sizeof(ch[sz]));
return sz;
}

void extend(int c) {
int p = lst;
int np = creat(val[p] + 1);
while(p && ch[p][c] == 0)
ch[p][c] = np, p = par[p];
if(!p) par[np] = root;
else {
int q = ch[p][c];
if(val[q] == val[p] + 1)
par[np] = q;
else {
int nq = creat(val[p] + 1);
memcpy(ch[nq], ch[q], sizeof(ch[q]));
par[nq] = par[q];
par[q] = nq;
par[np] = nq;
while(p && ch[p][c] == q)
ch[p][c] = nq, p = par[p];
}
}
lst = np;
}

int b[mnx << 1], d[mnx];

char s[mnx];
int n, len;

int main() {
while(scanf("%d", &n) != EOF) {
sz = 0;
root = lst = creat(0);
len = 0;
for(int i = 1; i <= n; ++i) {
scanf("%s", s);
len += strlen(s);
for(int j = 0; s[j]; ++j)
extend(s[j] - '0');
extend(10);
++len;
}
memset(d, 0, sizeof(d));
for(int i = 1; i <= sz; ++i) ++d[val[i]];
for(int i = 1; i <= len; ++i) d[i] += d[i - 1];
for(int i = 1; i <= sz; ++i) b[d[val[i]]--] = i;
memset(cnt, 0, sizeof(cnt));
memset(sum, 0, sizeof(sum));
cnt[1] = 1;
for(int i = 1; i <= sz; ++i) {
int u = b[i];
for(int j = 0; j < 10; ++j) {
if(u == 1 && j == 0) continue; //去除前导0的情况。
if(ch[u][j] == 0) continue;
int v = ch[u][j];
int add = (sum[u] * 10 + j * cnt[u]) % mod;
sum[v] = (sum[v] + add) % mod;
cnt[v] += cnt[u];
cnt[v] %= mod;
}
}
int ans = 0;
for(int i = 1; i <= sz; ++i)
ans = (ans + sum[i]) % mod;
printf("%d\n", ans);
}
return 0;
}