HDU

题目：给你n个由0~9组成的字符串，求不同的子串表示的十进制数的和，对2012取模

思路：将这n个串用10连接起来，建立一个SAM，然后跑拓扑序，注意含有前导0的数不要计算否则会重复计算

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int MOD=2012;const int MAXN = 300050, SIZE = 11;struct SAM {    int len[MAXN], link[MAXN], next[MAXN][SIZE];    int total, last;    inline int newNode(int L) {        len[++total] = L; link[total] = 0;        for(int i = 0; i < SIZE; ++i) next[total][i] = 0;        return total;    }    inline void Add(int c) {        int i, p = last, cur = newNode(len[last] + 1);        for(; p && !next[p][c]; p = link[p]) next[p][c] = cur;        if(!p) link[cur] = 1;//令其指向初始状态        else {            int q = next[p][c];            if(len[q] == len[p] + 1) link[cur] = q;            else {//>                int clone = newNode(len[p] + 1);                for(i = 0; i < SIZE; ++i) next[clone][i] = next[q][i];                link[clone] = link[q];                link[q] = link[cur] = clone;for(; p && next[p][c] == q; p = link[p]) next[p][c] = clone;            }        }        last = cur;    }    void Init () {//根节点是1        total = 0;        last = newNode(0);    }}sam;char str[MAXN];int num[MAXN],idx[MAXN];int sum[MAXN],tim[MAXN];int main(){    int n;    while(~scanf("%d",&n)){        sam.Init();        while(n--){            scanf("%s",str);            for(int i=0;str[i]!='\0';i++)                sam.Add(str[i]-'0');            sam.Add(10);        }        mm(num,0);        for(int i=1;i<=sam.total;i++) num[sam.len[i]]++;        for(int i=1;i<=sam.total;i++) num[i]+=num[i-1];        for(int i=1;i<=sam.total;i++) idx[num[sam.len[i]]--]=i;        mm(sum,0);        mm(tim,0);        int ans=0;        tim[1]=1;        for(int i=1;i<=sam.total;i++){            int p=idx[i];            ans=(ans+sum[p])%MOD;            for(int j=0;j<10;j++){                if(p==1&&j==0)//去除有前导0的数                    continue;                if(sam.next[p][j]){                    int np=sam.next[p][j];                    sum[np]=(sum[np]+10*sum[p]%MOD+tim[p]*j%MOD)%MOD;                    tim[np]=(tim[np]+tim[p])%MOD;                }            }        }        printf("%d\n",ans);    }    return 0;}