SRM 595 DIV2 1000

数位DP的感觉，但是跟模版不是一个套路的，看的题解，代码好理解，但是确实难想。
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 #define LL long long
 6 LL dp[31][2][2][2];
 7 int a[31],b[31],c[31];
 8 void fun(int *p,int x)
 9 {
10     int i;
11     for(i = 0; i <= 30; i ++)
12     {
13         if(x&(1<<i))
14             p[i] = 1;
15         else
16             p[i] = 0;
17     }
18 }
19 LL dfs(int pos,int ta,int tb,int tc)
20 {
21     if(pos == -1)
22     return 1;
23     LL & res = dp[pos][ta][tb][tc];
24     int x,y,z;
25     if(res == -1)
26     {
27         res = 0;
28         for(x = 0; x < 2; x ++)
29         {
30             for(y = 0; y < 2; y ++)
31             {
32                 z = x^y;
33                 if((!ta||(x <= a[pos]))&&(!tb||(y <= b[pos]))&&(!tc||(z <= c[pos])))
34                 {
35                     res += dfs(pos-1,ta&&(x == a[pos]),tb&&(y == b[pos]),tc&&(z == c[pos]));
36                 }
37             }
38         }
39     }
40     return res;
41 }
42 class LittleElephantAndXor
43 {
44 public :
45     LL getNumber(int A, int B, int C)
46     {
47         memset(dp,-1,sizeof(dp));
48         fun(a,A);
49         fun(b,B);
50         fun(c,C);
51         return dfs(30,1,1,1);
52     }
53 };