分析 :
这道题对于单个串的用 SAM 然后想想怎么维护就行了
但是多个串下、可以先将所有的串用一个不在字符集( 这道题的字符集是 '0' ~ '9' )
链接起来、建立后缀自动机之后
在统计答案的时候直接忽略掉不合法的边集的状态转移即可
#include<bits/stdc++.h> #define LL long long #define ULL unsigned long long #define scl(i) scanf("%lld", &i) #define scll(i, j) scanf("%lld %lld", &i, &j) #define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k) #define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l) #define scs(i) scanf("%s", i) #define sci(i) scanf("%d", &i) #define scd(i) scanf("%lf", &i) #define scIl(i) scanf("%I64d", &i) #define scii(i, j) scanf("%d %d", &i, &j) #define scdd(i, j) scanf("%lf %lf", &i, &j) #define scIll(i, j) scanf("%I64d %I64d", &i, &j) #define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k) #define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k) #define scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k) #define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l) #define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l) #define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l) #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define lowbit(i) (i & (-i)) #define mem(i, j) memset(i, j, sizeof(i)) #define fir first #define sec second #define VI vector<int> #define ins(i) insert(i) #define pb(i) push_back(i) #define pii pair<int, int> #define VL vector<long long> #define mk(i, j) make_pair(i, j) #define all(i) i.begin(), i.end() #define pll pair<long long, long long> #define _TIME 0 #define _INPUT 0 #define _OUTPUT 0 clock_t START, END; void __stTIME(); void __enTIME(); void __IOPUT(); using namespace std; const LL mod = (LL)1e9 + 7; struct SAM { static const int MAXN = ((int)1e6 + 10)<<1;//大小为字符串长度两倍 static const int LetterSize = 26; int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN]; int tpSum[MAXN], tp[MAXN], cnt[MAXN]; //tpSum,tp用于拓扑排序，tp为排序后的数组 void init(void){ last = tot = 1; len[1] = 0; memset( ch[1], 0, sizeof ch[1]); } void add( int x){ int p = last, np = last = ++tot; len[np] = len[p] + 1, cnt[last] = 1; memset( ch[np], 0, sizeof ch[np]); while( p && !ch[p][x]) ch[p][x] = np, p = fa[p]; if( p == 0) fa[np] = 1; else { int q = ch[p][x]; if( len[q] == len[p] + 1) fa[np] = q; else { int nq = ++tot; memcpy( ch[nq], ch[q], sizeof ch[q]); len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq; while( p && ch[p][x] == q) ch[p][x] = nq, p = fa[p]; } } } void toposort(void){ for(int i = 1; i <= len[last]; i++) tpSum[i] = 0; for(int i = 1; i <= tot; i++) tpSum[len[i]]++; for(int i = 1; i <= len[last]; i++) tpSum[i] += tpSum[i-1]; for(int i = 1; i <= tot; i++) tp[tpSum[len[i]]--] = i; //for(int i = tot; i; i--) cnt[fa[tp[i]]] += cnt[tp[i]]; } LL sum[MAXN]; LL GetAns(){ mem(cnt, 0); cnt[1] = 1; for(int i=1; i<=tot; i++){ int cur = tp[i]; if(!cnt[cur]) continue; for(int c=0; c<10; c++){ int nxt = ch[cur][c]; if(nxt){ cnt[nxt] = (cnt[nxt] + cnt[cur]) % mod; sum[nxt] = (sum[nxt] + (sum[cur] * 10 % mod + 1LL * c * cnt[cur] % mod) % mod) %mod; } } } LL ret = 0; for(int i=1; i<=tot; i++) ret = (ret + sum[i]) % mod; return ret; } } sam; int n; char str[(int)1e6 + 10]; int main(void){__stTIME();__IOPUT(); sci(n); sam.init(); for(int i=0; i<n; i++){ scs(str); int len = strlen(str); for(int j=0; j<len; j++) sam.add(str[j] - '0'); if(i != n-1) sam.add(10); } sam.toposort(); printf("%lld ", sam.GetAns()); __enTIME();return 0;} void __stTIME() { #if _TIME START = clock(); #endif } void __enTIME() { #if _TIME END = clock(); cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl; #endif } void __IOPUT() { #if _INPUT freopen("in.txt", "r", stdin); #endif #if _OUTPUT freopen("out.txt", "w", stdout); #endif }