牛客练习赛33 E. tokitsukaze and Similar String （字符串哈希）

题目链接：https://ac.nowcoder.com/acm/contest/308/E

题意：中文题 见链接

题解：哈希预处理（三哈希模板）

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define ull unsigned long long
 #define mst(a,b) memset((a),(b),sizeof(a))
 #define mp(a,b) make_pair(a,b)
 #define pi acos(-1)
 #define pii pair<int,int>
 #define pb push_back
 const int INF = 0x3f3f3f3f;
 const double eps = 1e-;
 const int maxn = 1e5 + ;
 const int maxm = 1e5 + ;
 const ll mod =  1e9 + ;

 ll P[] = {,,};
 ll MO[] = {mod,mod + ,};
 ll T[maxn][];
 char s[maxn];

 struct node {
     ll has[];
 }a[maxn][];

 int main() {
 #ifdef local
     freopen("data.txt", "r", stdin);
 //    freopen("data.txt", "w", stdout);
 #endif
     for(int i = ; i < ; i++) T[][i] = ;
     for(int i = ; i < maxn; i++)
         for(int j = ; j < ; j++) T[i][j] = T[i - ][j] * P[j] % MO[j];
     int n;
     scanf("%d",&n);
     scanf("%s",s + );
     for(int i = ; i < ; i++)
         for(int j = ; j < ; j++) a[][j].has[i] = ;
     for(int j = ; j < ; j++) {
         for(int i = ; i <= n; i++) {
             s[i]++;
             if(s[i] > 'z') s[i] -= ;
             for(int k = ; k < ; k++) a[i][j].has[k] = (a[i - ][j].has[k] * P[k] % MO[k] + (s[i] - 'a')) % MO[k];
         }
     }
     int q;
     scanf("%d",&q);
     while(q--) {
         int x,y,len;
         scanf("%d%d%d",&x,&y,&len);
         ll ned[],now[];
         for(int i = ; i < ; i++) {
             ned[i] = a[y + len - ][].has[i] - a[y - ][].has[i] * T[len][i] % MO[i];
             if(ned[i] < ) ned[i] += MO[i];
         }
         int ans = 1e9;
         for(int i = ; i < ; i++) {
             bool flag = true;
             for(int j = ; j < ; j++) {
                 now[j] = a[x + len - ][i].has[j] - a[x - ][i].has[j] * T[len][j] % MO[j];
                 if(now[j] < ) now[j] += MO[j];
                 if(now[j] != ned[j]) {
                     flag = false;
                     break;
                 }
             }
             if(flag)
                 ans = min(ans,min(i,  - i));
         }
         if(ans == 1e9) ans = -;
         printf("%d\n",ans);
     }
     return ;
 }