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。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define LL long long
LL dp[31][2][2][2];
int a[31],b[31],c[31];
void fun(int *p,int x)
{
    int i;
    for(i = 0; i <= 30; i ++)
    {
        if(x&(1<<i))
            p[i] = 1;
        else
            p[i] = 0;
    }
}
LL dfs(int pos,int ta,int tb,int tc)
{
    if(pos == -1)
    return 1;
    LL & res = dp[pos][ta][tb][tc];
    int x,y,z;
    if(res == -1)
    {
        res = 0;
        for(x = 0; x < 2; x ++)
        {
            for(y = 0; y < 2; y ++)
            {
                z = x^y;
                if((!ta||(x <= a[pos]))&&(!tb||(y <= b[pos]))&&(!tc||(z <= c[pos])))
                {
                    res += dfs(pos-1,ta&&(x == a[pos]),tb&&(y == b[pos]),tc&&(z == c[pos]));
                }
            }
        }
    }
    return res;
}
class LittleElephantAndXor
{
public :
    LL getNumber(int A, int B, int C)
    {
        memset(dp,-1,sizeof(dp));
        fun(a,A);
        fun(b,B);
        fun(c,C);
        return dfs(30,1,1,1);
    }
};