hdu-5661 Claris and XOR(贪心)

题目链接：

Claris and XOR

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

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

 

Input

The first line contains an integer 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(1≤a,b,c,d≤1018). 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 y.

 

Sample Input

2

1 2 3 4

5 7 13 15

 

Sample Output

6

11

 

题意：

 

问a<=x<=b,c<=y<=d;最大的x^y的值是多少;

 

思路：

 

贪心,从高位到低位,能取到1的取1不能的取0,同时更新a,b,c,d的值;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e18;
const int N=1e5+10;
const int maxn=5e3+4;
const double eps=1e-12;

LL a,b,c,d;

int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    read(T);
    while(T--)
    {
        read(a);read(b);read(c);read(d);
        LL ans=0;
        for(int i=63;i>=0;i--)
        {
            int fa=((a>>i)&1),fb=((b>>i)&1),fc=((c>>i)&1),fd=((d>>i)&1);
            if(fa!=fb&&fc!=fd){ans|=(1LL<<(i+1))-1;break;}
            else if(fa!=fb)
            {
                ans|=(1LL<<i);
                if(fc)
                {
                    b=(1LL<<i)-1;
                    c-=(1LL<<i);
                    d-=(1LL<<i);
                }
                else 
                {
                    a=0;
                    b-=(1LL<<i);
                }
            }
            else if(fc!=fd)
            {
                ans|=(1LL<<i);
                if(fa)
                {               
                    d=(1LL<<i)-1;
                    a-=(1LL<<i);
                    b-=(1LL<<i);
                }
                else 
                {
                    c=0;
                    d-=(1LL<<i);
                }
            }
            else
            {
                if(fa==fc)
                {
                    if(fa)
                    {
                        LL temp=(1LL<<i);
                        a-=temp;
                        b-=temp;
                        c-=temp;
                        d-=temp;
                    }
                }
                else 
                {
                    ans|=(1LL<<i);
                    if(fa)
                    {
                        a-=(1LL<<i);
                        b-=(1LL<<i);
                    }
                    else 
                    {
                        c-=(1LL<<i);
                        d-=(1LL<<i);
                    }
                }
            }
        }
        print(ans);
    }
    return 0;
}