Forethought Future Cup - Elimination Round

Posted 2021-02-17 gloid

　　A：签到。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
	while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n;char s[100];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	scanf("%s",s+1);n=strlen(s+1);int cnt=0;
	for (int i=1;i<=n;i++) if (s[i]==‘a‘) cnt++;
	for (int i=n;i>=1;i--) if (cnt>i/2) {cout<<i;break;} 
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：显然可能的划分位置只有一个，找到后检验即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
	while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n;char s[N],a[N],b[N];
signed main()
{
	scanf("%s",s+1);n=strlen(s+1);int cnt=0;
	int cnt2=0;for (int i=1;i<=n;i++) if (s[i]==‘a‘) cnt2++;
	for (int i=1;i<=n;i++)
	{
		if (s[i]==‘a‘) cnt++;
		if (cnt==cnt2&&i-cnt==n-i)
		{
			int t=0;
			for (int j=1;j<=i;j++) if (s[j]!=‘a‘) a[++t]=s[j];
			t=0;
			for (int j=i+1;j<=n;j++) b[++t]=s[j];
			for (int j=1;j<=t;j++) if (a[j]!=b[j]) {cout<<":(";return 0;}
			for (int j=1;j<=i;j++) putchar(s[j]);return 0;
		}
	}
	cout<<":(";
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：二进制分组。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
	while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int T,n,a[110],b[110];
signed main()
{
	T=read();
	while (T--)
	{
		n=read();
		int ans=0;
		for (int i=7;i>=0;i--)
		{
			int u=0,v=0;
			for (int j=1;j<=n;j++)
			if (j&(1<<i)) a[++u]=j;
			else b[++v]=j;
			if (u&&v)
			{
				cout<<u<<‘ ‘<<v<<‘ ‘;
				for (int j=1;j<=u;j++) cout<<a[j]<<‘ ‘;
				for (int j=1;j<=v;j++) cout<<b[j]<<‘ ‘;
				cout<<endl;
				ans=max(ans,read());
			}
		}
		cout<<-1<<‘ ‘<<ans<<endl;
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：不考虑范围限制的话显然能到达所有gcd(a,b)的倍数。注意到对于不小于a+b的gcd(a,b)的倍数，该点一定可以为到达该点时所经过的最大值。那么对于<a+b的数暴力跑dij，>=a+b的数等差数列求和即可。然后因为inf设小调了一年。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define M 1000000
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
	while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,a,b,d[1000010];
bool flag[1000010];
struct data
{
	int x,d;
	bool operator <(const data&a) const
	{
		return d>a.d;
	}
};
priority_queue<data> q;
void dij()
{
	for (int i=1;i<=M;i++) d[i]=1000000000;d[0]=0;
	q.push((data){0,0});
	for (;;)
	{
		while (!q.empty()&&flag[q.top().x]) q.pop();
		if (q.empty()) break;
		data x=q.top();q.pop();
		flag[x.x]=1;
		if (x.x>=b&&d[x.x-b]>x.d)
		{
			d[x.x-b]=x.d;
			q.push((data){x.x-b,d[x.x-b]});
		}
		if (x.x+a<=M&&d[x.x+a]>max(x.x+a,x.d))
		{
			d[x.x+a]=max(x.x+a,x.d);
			q.push((data){x.x+a,d[x.x+a]});
		}
	}
}
signed main()
{
	n=read(),a=read(),b=read();
	dij();
	ll ans=0;
	for (int i=0;i<=min(n,M);i++) ans+=max(0,n-d[i]+1);
	int u=gcd(a,b),cnt=n/u-M/u;
	if (cnt>0)
	{
		ans+=1ll*cnt*(n+1);
		int l,r;
		for (l=M+1;l<=n;l++) if (l%u==0) break;
		for (r=n;r>M;r--) if (r%u==0) break;
		ans-=1ll*(l+r)*cnt/2;
	}
	//??i (i=1000001~n,i%u==0)
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　EFGH：占坑待填。