hdu 4810 Wall Painting (组合数学+二进制)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <queue>
 5 #include <cstring>
 6 #include <map>
 7 #include <cstdlib>
 8 #include <algorithm>
 9 #define LL __int64
10 #define INF 0x3f3f3f3f
11 const int maxn = 1000+10;
12 const int mo = 1000000+3;
13 using namespace std;
14 LL c[maxn][maxn], a[35], p[35];  //注意要用LL，中间计算会超int
15 void init()
16 {
17     int i, j;
18     memset(c, 0, sizeof(c));
19     for(i = 0; i < maxn-5; i++)
20     c[i][i] = c[i][0] = 1;
21     for(i = 1; i < maxn-5; i++)
22     for(j = 1; j < i; j++)
23     c[i][j] = (c[i-1][j-1] + c[i-1][j])%mo;
24 
25     p[0] = 1;
26     for(i = 1; i < 32; i++)
27     p[i] = (2*p[i-1])%mo;
28 }
29 void cal(int x)
30 {
31     int cnt = 0;
32     while(x)
33     {
34         if(x%2)
35         a[cnt] ++;
36         x /= 2;
37         cnt ++;   //这个不要放到上面数组中，不然会在数组的下标多加一个
38     }
39 }
40 int main()
41 {
42     int n, i, j, k, x;
43     LL tmp, ans;
44     init();
45     while(~scanf("%d", &n))
46     {
47         memset(a, 0, sizeof(a));
48         for(i = 0; i < n; i++)
49         {
50             scanf("%d", &x);
51             cal(x);
52         }
53         for(i = 1; i <= n; i++)
54         {
55             ans = 0;
56             for(j = 0; j < 32; j++)
57             {
58                 for(k = 1; k <= i; k += 2)
59                 {
60                     tmp = ((LL)(p[j]*c[a[j]][k]*c[n-a[j]][i-k]))%mo; //该位的数值 * 该位为1的个数选奇数个 * 为0的个数选空着的个数。
61                      ans += tmp;
62                     ans %= mo;
63                 }
64             }
65             if(i == n)
66             printf("%I64d\n", ans);
67             else
68             printf("%I64d ", ans);
69         }
70     }
71     return 0;
72 }