UVALive 6507 Passwords

Passwords

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 6507
64-bit integer IO format: %lld      Java class name: Main

解题：好高深的hash啊

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 205;
 5 const int base = 233;
 6 const int mod = 1e9+7;
 7 char str[maxn][20010];
 8 LL hs[maxn][20010],B[20010];
 9 int n;
10 unordered_map<LL,int>ump;
11 void init(){
12     B[0] = 1;
13     for(int i = 1; i < 20010; ++i)
14         B[i] = B[i-1]*base%mod;
15 }
16 LL calc(int i,int L,int R){
17     return ((hs[i][R] - hs[i][L-1]*B[R - L + 1])%mod + mod)%mod;
18 }
19 void solve(){
20     ump.clear();
21     for(int i = 0; i < n; ++i){
22         for(int j = 1,len = strlen(str[i] + 1); j <= len; ++j){
23             hs[i][j] = (hs[i][j-1]*base + str[i][j])%mod;
24             ump[hs[i][j]]++;
25         }
26     }
27     int a = 0,b = 0;
28     for(int i = 0; i < n; ++i){
29         int len = strlen(str[i] + 1);
30         for(int j = 1; j <= len; ++j) --ump[hs[i][j]];
31         for(int j = 1; j <= len; ++j){
32             LL suffix = calc(i,len - j + 1,len);
33             if(!ump[suffix]) continue;
34             int x = 1,y = 1;
35             LL prefix = suffix;
36             for(int k = 2; k*j <= len; ++k){
37                 LL suffix2 = calc(i,len - j*k + 1,len);
38                 prefix = (prefix*B[j] + suffix)%mod;
39                 if(suffix2 != prefix) break;
40                 if(ump[prefix]) x = k;
41                 y = k;
42             }
43             if(x == y && x == 1) continue;
44             if(x == y) --x;
45             if(j*(x + y) > a + b){
46                 a = x*j;
47                 b = y*j;
48             }
49         }
50         for(int j = 1; j <= len; ++j) ump[hs[i][j]]++;
51     }
52     printf("%d %d\n",a,b);
53 }
54 int main(){
55     int kase;
56     init();
57     scanf("%d",&kase);
58     while(kase--){
59         scanf("%d",&n);
60         for(int i = 0; i < n; ++i)
61             scanf("%s",str[i] + 1);
62         solve();
63     }
64     return 0;
65 }
66 /*
67 2
68 3
69 abcabe
70 defg
71 bcabab
72 */