C. XOR-distance

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-28

48.C. XOR-distance2024-02-28

49.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

题解

既然是按位异或，那么我们尝试在二进制视角下考虑问题

我们发现，当两个数的同一位置上都有1的时候，这个1消与不消对结果没有影响，而这个位置上其中一个有一，另一个没有一时，我们可以通过x来转移一
所以在ab俩个数出现第一个10情况不同的位置之后，我们要尽可能地把剩余的一尽可能地移到较小的数
并且高位一加成效果远大于地位一，所以我们贪心移一

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
    ll t;
    cin>>t;
    while(t--)
    {
        ll a,b,r;
        cin>>a>>b>>r;
        ll x=0,i=log2(max(a,b)),len=log2(r);
        for(;i>=1;i--)//找第一个不同10的位置
        {

            if(((a>>i)&1)==1&&((b>>i)&1)==0)//判断一个位置上是一还是零的方法是右移该位置至第0位，然后与1
            {
                for(ll j=min(i-1,len);j>=0;j--)
                {
                    ll dif=(1LL<<j);
                    if(r-dif>=0&&(a&dif)!=0&&(b&dif)==0)
                    {
                        a^=dif;
                        b^=dif;
                        r-=dif;
                    }
                }
                break;
            }
            else if(((a>>i)&1)==0&&((b>>i)&1)==1)
            {
                for(ll j=min(i-1,len);j>=0;j--)
                {
                    ll dif=(1LL<<j);//这里一定要1LL！不然会溢出错误
                    if(r-dif>=0&&(a&dif)==0&&(b&dif)!=0)//对于与的判断是不等于零，因为与的结果实际上是该位置的2的次方！！
                    {
                        a^=dif;
                        b^=dif;
                        r-=dif;
                    }
                }
                break;
            }
        }
        cout<<abs(a-b)<<endl;
    }
    return 0;
}