Codeforces Round #570 Div. 3

　　A：暴力从小到大枚举判断。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
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;
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	int n=read();
	for (int i=n;i<=n+10000;i++)
	{
		int s=0,x=i;
		while (x) s+=x%10,x/=10;
		if (s%4==0) {cout<<i;return 0;}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：显然max-min<=2k时有解，最优解为min+k。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
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 q,n,k,a[N];
signed main()
{
	q=read();
	while (q--)
	{
		n=read(),k=read();
		for (int i=1;i<=n;i++) a[i]=read();
		int mn=inf,mx=0;
		for (int i=1;i<=n;i++) mn=min(mn,a[i]),mx=max(mx,a[i]);
		if (mx-k<=mn+k) cout<<mn+k<<endl;
		else cout<<-1<<endl; 
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：一点脑子都不想动可以二分答案。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
#define int 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 q,k,n,a,b;
signed main()
{
	q=read();
	while (q--)
	{
		k=read()-1,n=read(),a=read(),b=read();
		if (b*n>k) printf("-1
");
		else
		{
			int l=0,r=n,ans=0;
			while (l<=r)
			{
				int mid=l+r>>1;
				if (a*mid+b*(n-mid)<=k) ans=mid,l=mid+1;
				else r=mid-1;
			}
			printf("%d
",ans);
		}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D、E：跳过。

　　F：当只要求选择两个数时，显然选择最大值及最大的不是其因子的数最优。因为假设次大值是最大值的因子，那么次大值<=最大值/2，如果不选择最大值，两个数的和一定不大于最大值。于是考虑枚举最大值，剩下的数按上述方法选择。可以使用set删去因子，复杂度O(nsqrt(n)log(n))但也能跑过。实际上sort后枚举最大值并暴力找去掉因子的最大数复杂度就是O(nsqrt(n))。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
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 q,n,a[N],b[N],t;
set<int> f;
void del(int x)
{
	f.erase(x);
	b[++t]=x;
}
signed main()
{
	q=read();
	while (q--)
	{
		n=read();
		for (int i=1;i<=n;i++) a[i]=read();f.clear();
		sort(a+1,a+n+1);n=unique(a+1,a+n+1)-a-1;
		int ans=0;
		for (int i=1;i<=n;i++) f.insert(a[i]);
		for (int i=n;i>=1;i--)
		{
			t=0;
			int s=a[i];f.erase(a[i]);
			for (int j=1;j*j<=a[i];j++)
			if (a[i]%j==0)
			{
				if (f.find(j)!=f.end()) del(j);
				if (f.find(a[i]/j)!=f.end()) del(a[i]/j);
			}
			if (!f.empty())
			{
				int x=*(--f.end());
				for (int j=1;j*j<=x;j++)
				if (x%j==0) 
				{
					if (f.find(j)!=f.end()) del(j);
					if (f.find(x/j)!=f.end()) del(x/j);
				}
				s+=x;
			}
			if (!f.empty()) s+=*(--f.end());
			ans=max(ans,s);
			for (int j=1;j<=t;j++) f.insert(b[j]);
		}
		printf("%d
",ans);
	}
	return 0;
	//NOTICE LONG LONG!!!!!
} 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
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 q,n,a[N];
signed main()
{
	q=read();
	while (q--)
	{
		n=read();
		for (int i=1;i<=n;i++) a[i]=read();
		sort(a+1,a+n+1);n=unique(a+1,a+n+1)-a-1;
		int ans=0;
		for (int i=n;i>=1;i--) 
		{
			int s=a[i];int x=0;
			for (int j=i-1;j>=1;j--)
			if (a[i]%a[j]) {x=j,s+=a[j];break;}
			for (int j=x-1;j>=1;j--)
			if (a[i]%a[j]&&a[x]%a[j]) {s+=a[j];break;}
			ans=max(ans,s);
		}
		printf("%d
",ans);
	}
	return 0;
	//NOTICE LONG LONG!!!!!
} 

　　G：求出每种糖果数量以及其中有多少个f_i=1。然后从大到小考虑礼物中的糖果数量，将糖果数量足够的糖果种类加入set，取出Σf_i最大的。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 200010
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 q,n;
struct data
{
	int x,y;
	bool operator <(const data&a) const
	{
		return x<a.x||x==a.x&&y<a.y;
	}
}a[N],c[N];
multiset<int> f;
signed main()
{
	q=read();
	while (q--)
	{
		n=read();
		for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read();
		sort(a+1,a+n+1);
		int m=0;
		for (int i=1;i<=n;i++)
		{
			int t=i;
			while (t<n&&a[t+1].x==a[i].x) t++;
			c[++m].x=t-i+1;c[m].y=0;
			for (int j=i;j<=t;j++) if (a[j].y==1) c[m].y++;
			i=t;
		}
		sort(c+1,c+m+1);reverse(c+1,c+m+1);
		int cur=0,ans=0,ans2=0;f.clear();
		for (int i=n;i>=1;i--)
		{
			while (cur<m&&c[cur+1].x>=i)
			{
				f.insert(c[++cur].y);
			}
			if (!f.empty())
			{
				ans+=i;
				ans2+=min(i,*(--f.end()));
				f.erase(--f.end());
			}
		}
		printf("%d %d
",ans,ans2);
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　H：建出序列自动机（对每个位置求出每个字符下一次出现位置），然后dp出长度为i的本质不同子序列数量即可。注意数量对k取min。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 110
#define int 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,f[N][N],nxt[N][26];
ll k,ans;
char s[N];
signed main()
{
	n=read();ll k;cin>>k;
	scanf("%s",s+1);
	ans=k*n;
	for (int i=0;i<26;i++) nxt[n+1][i]=n+1;
	for (int i=n;i>=1;i--)
	{
		for (int j=0;j<26;j++) nxt[i][j]=nxt[i+1][j];
		
		nxt[i][s[i]-'a']=i;
	}
	for (int i=0;i<=n;i++) f[i][1]=1;
	for (int i=2;i<=n+1;i++)
		for (int j=n;j>=0;j--)
		{
			for (int k=0;k<26;k++)
			f[j][i]+=f[nxt[j+1][k]][i-1];
			f[j][i]=min(f[j][i],k);
		}
	for (int i=n+1;i>=1;i--)
	{
		if (k>f[0][i]) ans-=(i-1)*f[0][i],k-=f[0][i];
		else {ans-=(i-1)*k;k=0;break;}
	}
	if (k) cout<<-1;else 
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　小号5。result：rank 2 rating +322