codeforces 722F - Cyclic Cipher

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 const int N = 1e5 + 10, E = 2e5 + 10;
  4 int len[N], firste[N], nexte[E], v[E], w[E];
  5 int n, lim, e, ans, m;
  6 void build(int x, int y, int z)
  7 {
  8     nexte[++e] = firste[x];
  9     firste[x] = e;
 10     v[e] = y;
 11     w[e] = z;
 12 }
 13 void exgcd(long long a,long long b,long long &d, long long &x, long long &y)
 14 {
 15     if(!b)
 16     {
 17         x = 1;
 18         y = 0;
 19         d = a;
 20     }
 21     else
 22     {
 23         exgcd(b, a % b, d, y, x);
 24         y -= x * (a / b);
 25     }
 26 }
 27 long long a[17][N], b[17][N];
 28 long long mul(long long aa, long long bb, long long mod)
 29 {
 30     return (aa * bb - (long long)(aa / (long double) mod * bb + 1e-3) * mod + mod) % mod ;
 31 }
 32 void solve(long long m1, long long b1, long long m2, long long b2, long long &m0, long long &b0)
 33 {
 34     if(m1 == -1 || m2 == -1)
 35     {
 36         m0 = b0 = -1;
 37         return;
 38     }
 39     long long d, x, y;
 40     exgcd(m1, m2, d, x, y);
 41     if((b2 - b1) % d != 0)
 42         m0 = b0 = -1;
 43     else
 44     {
 45         long long t = m2 / d;
 46         //x = (x % t * ((b2 - b1) / d % t) % t + t) % t;
 47         x = (mul(x, (b2 - b1) / d, t) + t) % t;
 48         m0 = m1 * (m2 / d);
 49         b0 = m1 * x + b1;
 50     }
 51 }
 52 int main()
 53 {
 54     scanf("%d%d", &n, &lim);
 55     int x;
 56     for(int i = 1; i <= n; ++i)
 57     {
 58         scanf("%d", &len[i]);
 59         for(int j = 0; j < len[i]; ++j)
 60         {
 61             scanf("%d", &x);
 62             build(x, i, j);
 63         }
 64     }
 65     int last;
 66     for(int u = 1; u <= lim; ++u)
 67     {
 68         ans = m = last = 0;
 69         for(int p = firste[u]; ; p = nexte[p])
 70         {
 71             if(p && (!m || v[p] == last - 1))
 72             {
 73                 ++m;
 74                 a[0][m] = len[v[p]];
 75                 b[0][m] = w[p];
 76                 last = v[p];
 77                 continue;
 78             }
 79             else
 80             {
 81                 if(m > ans)
 82                 {
 83                     int top = 0;
 84                     for(int i = 1; (1 << i) <= m; ++i)
 85                     {
 86                         top = i;
 87                         for(int j = 1; j + (1 << i) - 1 <= m; ++j)
 88                             solve(a[i - 1][j], b[i - 1][j], a[i - 1][j + (1 << (i - 1))], 
 89                             b[i - 1][j + (1 << (i - 1))], a[i][j], b[i][j]);
 90                     }
 91                     for(int j = 1; m - j + 1 > ans; ++j)
 92                     {
 93                         int k = j + 1;
 94                         long long nowa = a[0][j], nowb = b[0][j], tmpa, tmpb;
 95                         for(int i = top; i >= 0; --i)
 96                             if(k + (1 << i) - 1 <= m)
 97                             {
 98                                 solve(nowa, nowb, a[i][k], b[i][k], tmpa, tmpb);
 99                                 if(tmpa != -1)
100                                 {
101                                     k += (1 << i);
102                                     nowa = tmpa;
103                                     nowb = tmpb;
104                                 }
105                             }
106                         ans = max(ans, k - j);
107                     }
108                 }
109                 if(!p)
110                     break;
111                 m = 1;
112                 a[0][m] = len[v[p]];
113                 b[0][m] = w[p];
114                 last = v[p];
115             }
116         }
117         printf("%d\n", ans);
118     }
119     return 0;
120 }