## problem

• $1 \le W \le A$
• $1 \le X \le B$
• $1 \le Y \le C$
• $1 \le Z \le D$
• $A \le B \le C \le D$
• $W \le X \le Y \le Z$
• $W \oplus X \oplus Y \oplus Z \ne 0$
• $\oplus$は排他的論理和

## solution

Use meet-in-middle technique. $O(N^2)$.

Let $A \le B \le C \le D$ and $W \le X \le Y \le Z$. Make $\operatorname{cnt}_l(s) = |\{ (y,z) \mid l \le y \le z, y \oplus z = s \}|$ and calculate $\rm{ans} = \Sigma_{1 \le x \le B} \Sigma_{1 \le w \le \min \{ x, A \}} (\rm{total} - \operatorname{cnt}_x(w \oplus x))$. You can construct $\operatorname{cnt}_{x+1}$ from $\operatorname{cnt}_x$ with $O(N)$.

## implementation

#include <iostream>
#include <vector>
#include <algorithm>
#include <array>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
typedef long long ll;
using namespace std;
int main() {
array<int,4> a; repeat (i,4) cin >> a[i];
sort(a.begin(), a.end());
vector<ll> cnt(pow(2,ceil(log2(a[3]+1))));
ll acc = 0;
repeat_from (y,1,a[2]+1) {
repeat_from (z,y,a[3]+1) {
cnt[y^z] += 1;
acc += 1;
}
}
ll ans = 0;
repeat_from (x,1,a[1]+1) {
repeat_from (w,1,min(a[0],x)+1) {
ans += acc - cnt[w^x];
}
int y = x;
repeat_from (z,x,a[3]+1) {
cnt[y^z] -= 1;
acc -= 1;
}
}
cout << ans << endl;
return 0;
}