E2. Game with Marbles (Hard Version)
1.C. Insert and Equalize2.C. Removal of Unattractive Pairs3.D. Jumping Through Segments4.E. Good Triples5.F. Shift and Reverse6.D. Yet Another Monster Fight7.A. Constructive Problems8.C. Game with Multiset9.A. Rating Increase10.B. Swap and Delete11.A. Problemsolving Log12.B. Preparing for the Contest13.C. Quests14.D. Three Activities
15.E2. Game with Marbles (Hard Version)
16.cf刷题有感17.A. Anonymous Informant18.A. Forked!19.B. Make Almost Equal With Mod20.C. Heavy Intervals21.D. Split Plus K22.A. 202323.B. Two Divisors24.C. Training Before the Olympiad25.D. Mathematical Problem26.F. Greetings27.C. Partitioning the Array28.G. Bicycles29.E. Eat the Chip30.G. Lights31.D. Array Repetition32.D. Berserk Monsters33.E. Increasing Subsequences34.D. Very Different Array35.G. Mischievous Shooter36.B. Plus-Minus Split37.B. A Balanced Problemset38.C. Did We Get Everything Covered39.D. Find the Different Ones!40.C. Grouping Increases41.D. Good Trip42.C. Physical Education Lesson43.E. Final Countdown44.D. Divisible Pairs45.G. Vlad and Trouble at MIT46.A. Brick Wall47.B. Minimize Inversions48.C. XOR-distance49.A. Moving Chips50.B. Monsters Attack!51.C. Find B52.D. Slimes53.C. Turtle Fingers: Count the Values of k54.D. Turtle Tenacity: Continual Mods55.D. Vlad and Division56.C. LR-remainders57.D. Lonely Mountain Dungeons导论,有点博弈论的感觉?
每个人轮流选一个大家都有的球,然后自己扣一个球,对方扣完。问女生剩下的球减去男生剩下的球,最大值是多少?
一些条件
1.初始每个人每种球都有
2.女生想使答案值大一点,男生想使答案值小一点,换句话说,女生想使
3.如果选择扣掉对面的球
总述
每个人选择的最优策略是选择
代码
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[200005]={0};
ll b[200005]={0};
struct unit
{
int v;
int id;
}sum[200005];
bool cmp(unit x,unit y)
{
return x.v>y.v;
}
int main()
{
ll t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
for(int i=1;i<=n;i++) cin>>a[i];
for(int i=1;i<=n;i++)
{
cin>>b[i];
sum[i].v=a[i]+b[i];//又要扣除,又要保留
sum[i].id=i;
}
sort(sum+1,sum+n+1,cmp);
ll who=0;
for(int i=1;i<=n;i++)
{
int index=sum[i].id;
if(who==0)
{
a[index]--;
b[index]=0;
}
else
{
a[index]=0;
b[index]--;
}
who=1-who;
}
ll ans=0;
for(int i=1;i<=n;i++)ans=ans+a[i]-b[i];
printf("%lld\n",ans);
}
return 0;
}
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