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hdu-5661 Claris and XOR(贪心)

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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 ab and cd. He wants to choose two integers x,y that satisfies axb and cyd, 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(1T10,000)——The number of the test cases.
For each test case, the only line contains four integers a,b,c,d(1a,b,c,d1018). 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(‘\n‘);
}
 
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;
}

  

 

hdu-5661 Claris and XOR(贪心)

原文:http://www.cnblogs.com/zhangchengc919/p/5759259.html

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