2022 Hangzhou Electric Multi-School 1st Game

A

B. Dragon slayer

• 题意

• 思路

• 代码

C. Backpack

• 题意
  Find what is the maximum XOR sum of the values ​​of all items that just fill the backpack,不存在输出-1.
• 思路
  Is a backpack model,Three dimensions can be considereddp,$dp[i,j,k]$表示前$i$个物品,异或和是$j$,The occupied volume is exactly $k$的方案是否存在.Obviously, the state transition is a simple choice or no choice, which can be transferred：$dp[i,j,k]=dp[i,j,k]|dp[i,j\oplus w,k-v]$.其中,$x=a\oplus b$,已知$x$和$a$求b,则有$b=a\oplus x$.这个代码如下,会TLE.

while(t--)
{
    
	scanf("%d%d",&n,&m);
	for(int i=0;i<=1024;i++) for(int j=0;j<=1024;j++) f[i][j]=0;
	f[0][0]=1;
	for(int i=1;i<=n;i++)
	{
    
		scanf("%d%d",&v,&w);
		f[w][v]=1;
		for(int j=0;j<1024;j++)
		{
    
			for(int k=v;k<=m;k++)
				f[j][k]|=f[j^w][k-v];
		}
	}
	int ans=-1;
	for(int i=0;i<1024;i++) if(f[i][m]) ans=i;
	cout<<ans<<endl;
}

于是考虑使用bitsetPress position,The layer of the cycle volume is optimized away,代码如下：

• 代码

#include<bits/stdc++.h>
using namespace std;
#define N 1050
int n,m,t,x,y;
bitset<N> f[N],g[N];
int main()
{
    
	cin>>t;
	while(t--)
	{
    
		scanf("%d%d",&n,&m);
		for(int j=0;j<1024;++j) f[j].reset();
		f[0][0]=1;
		for(int i=1;i<=n;++i)
		{
    
			cin>>x>>y;
			for(int j=0;j<1024;++j)
			{
    
				g[j]=f[j];
				g[j]<<=x;
			}
			for(int j=0;j<1024;++j)
			{
    
				f[j]|=g[j^y];
			}
		}
		int ans=-1;
		for(int i=0;i<1024;i++) if(f[i][m]) ans=i;
		cout<<ans<<endl;
	}
	return 0;
}

D

E

F

G

H

I. Laser

• 题意
  给定nThe coordinates of the enemy$(x_i,y_i)$,There is now a laser transmitter,他可以向8个方向（东、南、西、北、西北、东北、西南、东南）Launch self-centered8条射线.Ask if there is a point that can be placed this laser transmitter,Makes him kill all enemies with one shot.
• 思路
  Assume any pointQon a horizontal line,Find a point arbitrarilyPnot on this horizontal line.Then the launch center must be at the pointQcentered“米”word and dotPon the three intersections of the straight lines.Then just enumerate those three intersections,$O(n)$Determine if you can kill all enemies.
  同理,Assume any pointQon a vertical line/主对角线上/副对角线上,Repeat the above operation4次即可,为了方便实现,We rotate the coordinates,Rotate all lines to horizontal in this case.
• 代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 500010
int t,n;
ll a[N],b[N],x[N],y[N];

bool work(ll xx,ll yy)
{
    
	for(int i=1;i<=n;i++) 
		if(!(x[i]==xx || y[i]==yy || yy-xx==y[i]-x[i] || yy+xx==y[i]+x[i]))
			return false;
	return true;
}

bool check()
{
    
	int p=0;
	for(int i=2;i<=n;i++)
	{
    
		if(x[i]==x[1]) continue;
		p=1;
		if(work(x[1],y[i])) return true; //Note that here is finding a point that is not on this line Do it at this point“米”The intersection of the word and the original determined line.
		if(work(x[1],y[i]-(x[i]-x[1]))) return true;
		if(work(x[1],y[i]+(x[i]-x[1]))) return true;
 		break;
	}
	if(p==0) return true; //Note the points that are specially judged on a line
	return false;
}

int main()
{
    
	cin>>t;
	while(t--)
	{
    
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%lld%lld",&a[i],&b[i]);
		bool is_ok=false;
		for(int i=1;i<=n;i++)x[i]=a[i],y[i]=b[i];
		if(check()) is_ok=true; 
		for(int i=1;i<=n;i++)x[i]=b[i],y[i]=a[i];
		if(check()) is_ok=true; 
		for(int i=1;i<=n;i++)x[i]=a[i]+b[i],y[i]=a[i]-b[i];
		if(check()) is_ok=true; 
		for(int i=1;i<=n;i++)x[i]=a[i]-b[i],y[i]=a[i]+b[i];
		if(check()) is_ok=true; 
		if(is_ok) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

J

K. Random

• 题意
  给定n个分布在$[0,1]$之间的数,现在有m次操作,每次操作有$1/2$The probability of removing the smallest of all numbers now,有$1/2$The probability of removing the current largest number among all numbers.Ask what the expectation is for the sum of the last remaining numbers（答案mod1e9+7）.
• 思路
  after each operationnThe probability of the number remaining at a position（A few more observations will reveal that it is$n-m$）,Finally add these probabilities together and multiply$1/2$（The expected value of each number is $1/2$）,Because of the modulo operation,Hence the need to use an inverse.
• 代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const ll mod =1e9+7;
ll n,m,t;
ll qmi(ll a, ll k, ll p)    
{
    
    ll res = 1;
    while (k)
    {
    
        if (k & 1) res = (ll)res * a % p;
        a = (ll)a * a % p;
        k >>= 1;
    }
    return res%p;
}

int main()
{
    
	cin>>t;

	while(t--)
	{
    
		cin>>n>>m;
		if(n==m) printf("0\n");
		else
		{
    
			if(m==0) printf("%lld\n",(n*qmi(2,mod-2,mod))%mod);
			else
			{
    
				printf("%lld\n",((n-m)%mod*qmi(2,mod-2,mod))%mod);
			}
		}
	}
	return 0;
}

L. Alice and Bob

• 题意
  There are many numbers on the blackboard,范围是0-n之间.Alice和Bob玩游戏,AliceDivide the numbers on the blackboard into two sets at a time.BobDelete one of the collections at a time,Subtract one from each number in the other set.当出现一个0时,Alice Win;When all numbers are removedBob Win.Both are extremely smart,现给定0-n每个数的个数,Ask who is the lastWin.
• 思路
  Observation found that there should be2个1,4个2,8个3…The first mover wins.Two large numbers can be turned into a small one.因此循环a[i]+=a[i+1]/2.最后判断a[0]是否大于1即可.
• 代码

#include<bits/stdc++.h>
using namespace std;
#define N 1000010
int t,a[N],n;
int main()
{
    
	cin>>t;
	while(t--)
	{
    
		scanf("%d",&n);
		for(int i=0;i<=n;i++) scanf("%d",&a[i]);
		for(int i=n;i>=1;i--) a[i-1]+=a[i]/2;
		if(a[0]>0) printf("Alice\n");
		else printf("Bob\n");
	}
	return 0;
}