題目連結：https://ac.nowcoder.com/acm/contest/308/E

題意：中文題 見連結

題解：雜湊預處理（三雜湊模板）

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define ull unsigned long long
 5 #define mst(a,b) memset((a),(b),sizeof(a))
 6 #define mp(a,b) make_pair(a,b)
 7 #define pi acos(-1)
 8 #define
 
 pii pair<int,int>
 9 #define pb push_back
10 const int INF = 0x3f3f3f3f;
11 const double eps = 1e-6;
12 const int maxn = 1e5 + 10;
13 const int maxm = 1e5 + 10;
14 const ll mod =  1e9 + 7;
15  
16 ll P[3] = {359,1039,2797};
17 ll MO[3] = {mod,mod + 2,1000000087};
18 ll T[maxn][3];
19 char s[maxn];
20  
21
 
 struct node {
22     ll has[3];
23 }a[maxn][27];
24  
25  
26 int main() {
27 #ifdef local
28     freopen("data.txt", "r", stdin);
29 //    freopen("data.txt", "w", stdout);
30 #endif
31     for(int i = 0; i < 3; i++) T[0][i] = 1;
32     for(int i = 1; i < maxn; i++)
33         for(int j = 0; j < 3
 
; j++) T[i][j] = T[i - 1][j] * P[j] % MO[j];
34     int n;
35     scanf("%d",&n);
36     scanf("%s",s + 1);
37     for(int i = 0; i < 3; i++)
38         for(int j = 0; j < 26; j++) a[0][j].has[i] = 1;
39     for(int j = 0; j < 26; j++) {
40         for(int i = 1; i <= n; i++) {
41             s[i]++;
42             if(s[i] > 'z') s[i] -= 26;
43             for(int k = 0; k < 3; k++) a[i][j].has[k] = (a[i - 1][j].has[k] * P[k] % MO[k] + (s[i] - 'a')) % MO[k];
44         }
45     }
46     int q;
47     scanf("%d",&q);
48     while(q--) {
49         int x,y,len;
50         scanf("%d%d%d",&x,&y,&len);
51         ll ned[3],now[3];
52         for(int i = 0; i < 3; i++) {
53             ned[i] = a[y + len - 1][0].has[i] - a[y - 1][0].has[i] * T[len][i] % MO[i];
54             if(ned[i] < 0) ned[i] += MO[i];
55         }
56         int ans = 1e9;
57         for(int i = 0; i < 26; i++) {
58             bool flag = true;
59             for(int j = 0; j < 3; j++) {
60                 now[j] = a[x + len - 1][i].has[j] - a[x - 1][i].has[j] * T[len][j] % MO[j];
61                 if(now[j] < 0) now[j] += MO[j];
62                 if(now[j] != ned[j]) {
63                     flag = false;
64                     break;
65                 }
66             }
67             if(flag)
68                 ans = min(ans,min(i, 26 - i));
69         }
70         if(ans == 1e9) ans = -1;
71         printf("%d\n",ans);
72     }
73     return 0;
74 }