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 };