Poj_2939 Flavius Joseph PHUs reloaded

In the internal competition, this question is sdut_2369. Map writes TLE. In other words, the OJ team members were sentenced to AC and me to TLE. Finally, the Captain concluded that this is a matter of character... Khan = _ = ///

After the game, I wrote it with hash. The key is to handle conflicts and write it with the chain method. Finally, let's take a look at how it looks like an adjacent table. Maybe you can change this question to a good topic.

Poj 1235 + MS, sdutoj 578 + MS. Our server is still very powerful.

Slag code:

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 1 #include <vector>
 2 #include <list>
 3 #include <map>
 4 #include <set>
 5 #include <queue>
 6 #include <deque>
 7 #include <stack>
 8 #include <bitset>
 9 #include <algorithm>
10 #include <functional>
11 #include <numeric>
12 #include <utility>
13 #include <sstream>
14 #include <iostream>
15 #include <iomanip>
16 #include <cstdio>
17 #include <cmath>
18 #include <cstdlib>
19 #include <ctime>
20 #include <cstring>
21 
22 using namespace std;
23 
24 const int MOD = 1000007;
25 typedef long long ll;
26 struct node {
27     long long n;
28     int cnt;
29     int next;
30 } mp[MOD];
31 
32 int head[MOD], t;
33 
34 void init() {
35     for(int i = 0; i < MOD; ++i)    head[i] = 0;
36     t = 1;
37 }
38 
39 void add(int u, long long v) {
40     mp[t].n = v; mp[t].cnt = 1;
41     mp[t].next = head[u]; head[u] = t++;
42 }
43 
44 int find(int t, long long x) {
45     int i;
46     for(i = head[t]; i; i = mp[i].next) {
47         if(mp[i].n == x) {
48             mp[i].cnt++;
49             return mp[i].cnt;
50         }
51     }
52     return 0;
53 }
54 
55 int main() {
56     //freopen("data.in", "r", stdin);
57 
58     ll x;
59     int n, a, b, u;
60     int cnt;
61     while(scanf("%d", &n), n) {
62         scanf("%d%d", &a, &b);
63         init();
64         x = b%n; cnt = 0;
65 
66         while(1) {
67             u = find(x%MOD, x);
68             if(u == 2) cnt++;
69             if(u == 3)    break;
70             if(!u) {
71                 add(x%MOD, x);
72             }
73             x = (((a*x)%n*x) + b)%n;
74         }
75         printf("%d\n", n - cnt);
76     }
77     return 0;
78 }