Claris and XOR

Problem Description

Claris loves bitwise operations very much, especially XOR, because it has many beautiful features. He gets four positive integers a,b,c,da,b,c,d that satisfies a\leq ba≤b and c\leq dc≤d. He wants to choose two integers x,yx,y that satisfies a\leq x\leq ba≤x≤b and c\leq y\leq dc≤y≤d, and maximize the value of x~XOR~yx XOR y. But he doesn't know how to do it, so please tell him the maximum value of x~XOR~yx XOR y.

Input

The first line contains an integer T\left(1\leq T\leq10,000\right)T(1≤T≤10,000)——The number of the test cases. For each test case, the only line contains four integers a,b,c,d\left(1\leq a,b,c,d\leq10^{18}\right)a,b,c,d(1≤a,b,c,d≤10​18​​). Between each two adjacent integers there is a white space separated.

Output

For each test case, the only line contains a integer that is the maximum value of x~XOR~yx XOR y.

Sample Input

2
1 2 3 4
5 7 13 15

Sample Output

Copy

6
11

Hint

In the first test case, when and only when x=2,y=4x=2,y=4, the value of x~XOR~yx XOR y is the maximum. In the second test case, when and only when x=5,y=14x=5,y=14 or x=6,y=13x=6,y=13, the value of x~XOR~yx XOR y is the maximum.

 

 

 

这游戏真难,之前写过一道类似的,就直接敲了...o(n)...超时....

 

 

TLE代码:

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long

int const MAX = 100000005;
int n;

struct Trie
{
    int root, tot, next[MAX][2], end[MAX];
    inline int node()
    {
        memset(next[tot], -1, sizeof(next[tot]));
        end[tot] = 0;
        return tot ++;
    }

    inline void Init()
    {
        tot = 0;
        root = node();
    }

    inline void insert(ll x)
    {
        int p = root;
        for(int i = 31; i >= 0; i--)
        {
            int ID = ((1 << i) & x) ? 1 : 0;
            if(next[p][ID] == -1)
                next[p][ID] = node();
            p = next[p][ID];
        }
        end[p] = x;
    }

    inline int search(int x)
    {
        int p = root;
        for(int i = 31; i >= 0; i--)
        {
            int ID = ((1 << i) & x) ? 1 : 0;
            if(ID == 0)
                p = next[p][1] != -1 ? next[p][1] : next[p][0];
            else
                p = next[p][0] != -1 ? next[p][0] : next[p][1];
        }
        return x ^ end[p];
    }

}trie;

int a[2],b[2];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n=4;
        int WTF = 0,ans=0, x;
        trie.Init();
        for(int i = 0; i < 2; i++)
        {
            scanf("%d", &a[i]);
        }
        for(int i = 0; i < 2; i++)
        {
            scanf("%d", &b[i]);
        }
        for(int i=a[0]; i<=a[1]; i++){
            //WTF=0;
            //trie.insert(1);
            //WTF = WTFx(WTF, trie.search(1));
            for(int j=b[0]; j<=b[1]; j++){
                trie.Init();
                trie.insert(i);
                WTF = max(WTF, trie.search(i));
                trie.insert(j);
                WTF = max(WTF, trie.search(j));
                ans=max(WTF,ans);
            }
        }    
        printf("%d\n", WTF);
    }
}