NOIP2014提高组解方程

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <queue>
 4 
 5 using namespace std;
 6 
 7 int mod[7 + 2] = { 19 , 101 , 11261 , 19997 , 22877 , 21893 , 14843 };
 8 int n , m , ans[1000000 + 2];
 9 char a[100 + 10][10000 + 10];
10 int i , j , k;
11 int power[1000000 + 2];
12 int ver[7][1000000 + 2];
13 queue < int > q;
14 int num;
15 
16 bool check( int x )
17 {
18     int i , j , flag = 0;
19     long long ans = 0;
20     power[0] = 1;
21     for( i = 1 ; i <= n ; i++ )
22          power[i] = power[i - 1] * x % mod[k];
23     for( i = 0 ; i <= n ; i++ )
24     {
25          if( a[i][0] != ‘-‘ )
26              ans += ver[k][i] * power[i];
27          else
28              ans -= ver[k][i] * power[i];
29          ans %= mod[k];
30     }
31     while( ans < 0 )
32         ans += mod[k];
33     return !( ans % mod[k] );
34 }
35 
36 int main()
37 {
38     scanf( "%d %d" , &n , &m );
39     for( i = 0 ; i <= n ; i++ )
40          scanf( "%s" , a[i] );  
41     for( k = 0 ; k < 7 ; k++ )
42         for( i = 0 ; i <= n ; i++ )
43             for( j = 0 ; a[i][j] ; j++ )
44             {
45                 if( a[i][j] == ‘-‘ )
46                     continue;
47                 ver[k][i] *= 10;
48                 ver[k][i] += a[i][j] - ‘0‘;
49                 ver[k][i] %= mod[k];
50             }
51     for( k = 0 ; k < 7 ; k++ )
52          for( i = 0 ; i < mod[k] && i <= m ; i++ )
53               if( check( i ) )
54                   for( j = i ; j <= m ; j += mod[k] )
55                        ans[j]++;
56     for( i = 1 ; i <= m ; i++ )
57          if( ans[i] == 7 )
58          {
59              num++;
60              q.push( i );
61          }
62     printf( "%d\n" , num );
63     while( !q.empty() )
64          printf( "%d\n" , q.front() ) , q.pop();
65     return 0;
66 }