A

O(nmk)dp即可，因为带了1/2的常数+2s所以很稳

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
using namespace std;

int a[801];
int p[801][801];
long long f[801][801];
int n,m,i,j,k,l,K,Y;

int main()
{
//	freopen("a.in","r",stdin);
	
	scanf("%d%d%d%d",&n,&m,&K,&Y);
	fo(i,1,n)
	scanf("%d",&a[i]);
	fo(i,1,n)
	{
		fo(j,1,m)
		{
			scanf("%d",&p[i][j]);
			
			if (j<Y)
			p[i][j]+=a[i]*j;
		}
	}
	
	memset(f,127,sizeof(f));
	f[0][0]=0;
	
	fo(i,0,n-1)
	{
		fo(j,0,K)
		if (f[i][j]<800000000000ll)
		{
			fd(k,min(K-j,m),0)
			f[i+1][j+k]=min(f[i+1][j+k],f[i][j]+p[i+1][k]);
		}
	}
	
	printf("%lld
",f[n][K]);
}

B

m=2的约瑟夫问题，给出一个人求n=1~n时剩下这个人的方案&最大的n

约瑟夫的O(n)递推显然过不了，但是~~随便打表~~可以发现答案长这样：

1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1...

随便算

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
using namespace std;

//int f[233];
long long n,m,s,S,ans1,ans2;

int main()
{
//	freopen("b.in","r",stdin);
	
	scanf("%lld%lld",&n,&m);
	
	if (!(m&1))
	{
		printf("0 0
");
		return 0;
	}
	
	S=0;s=1;
	while (n)
	{
		if (n>=s)
		{
			n-=s;
			if (m<=s*2-1)
			++ans1,ans2=S+(m+1)/2;
		}
		else
		{
			if (m<=n*2-1)
			++ans1,ans2=S+(m+1)/2;
			
			n=0;
		}
		
		S+=s;
		s=s*2;
	}
	
	printf("%lld %lld
",ans1,ans2);
	return 0;
	
//	f[1]=0;
//	fo(i,2,100)
//	f[i]=(f[i-1]+2)%i;
//	
//	fo(i,1,100)
//	cout<<f[i]+1<<" ";
//	if (!f[i])
//	cout<<f[i-1]+1<<" ";
}

C

在普通循环同构的基础上加上了对角互换

~~懒得画~~

假设有一条直线穿过圆，并且保证直线上方的数<=对应的直线下方的数

这样解决了对角互换的条件

然后将圆旋转，可以发现上下两部分同时旋转了

于是可以把一对对应点看成一个元素，那么变成元素种数为m(m+1)/2，环的大小为n的普通轮换问题

直接O(n)枚举会挂，所以枚举gcd，再乘上phi(n/gcd)即可

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 19260817
#define Mod 19260815
using namespace std;

int f[mod+1];
int p[1226565];
int T,i,j,k,l,len,s;
long long n,m,ans;

void init()
{
	int i,j;
	
	fo(i,1,mod) f[i]=i;
	
	fo(i,2,mod)
	{
		if (f[i]==i)
		{
			--f[i];
			p[++len]=i;
		}
		
		fo(j,1,len)
		if ((long long)i*p[j]<=mod)
		{
			if (!(i%p[j]))
			{
				f[i*p[j]]=f[i]*p[j];
				break;
			}
			else
			f[i*p[j]]=f[i]*(p[j]-1);
		}
		else
		break;
	}
}

long long qpower(long long a,int b)
{
	long long ans=1;
	
	while (b)
	{
		if (b&1)
		ans=ans*a%mod;
		
		a=a*a%mod;
		b>>=1;
	}
	
	return ans;
}

int gcd(int n,int m)
{
	int r=n%m;
	
	while (r)
	{
		n=m;
		m=r;
		r=n%m;
	}
	
	return m;
}

int main()
{
//	freopen("c.in","r",stdin);
	
	init();
	
	scanf("%d",&T);
	for (;T;--T)
	{
		scanf("%lld%lld",&n,&m);
		m=m*(m+1)/2%mod;
		
		s=floor(sqrt(n));
		
		ans=0;
		fo(i,1,s)
		if (!(n%i))
		{
			ans=(ans+qpower(m,i)*f[n/i]%mod)%mod;
			if (i*i!=n)
			ans=(ans+qpower(m,n/i)*f[i]%mod)%mod;
		}
//		ans=(ans+qpower(m,gcd(n,i)))%mod;
		ans=ans*qpower(n,Mod)%mod;
		
		printf("%lld
",ans);
	}
}

D

sb题

把平面旋转45°再扩大(sqrt{2})倍（即(x,y)-->(x+y,y-x)），变成D*D的矩形操作

排序+扫描线

注意边界不能减

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define low(x) (x&-(x))
#define N 200001
using namespace std;

struct type{
	int x,y,s;
} a[200001];
int tr[4*N+1];
int tot,n,D,L,i,j,k,l,x,y,sum;
long long ans;

bool cmp(type a,type b)
{
	return a.x<b.x || a.x==b.x && a.s<b.s;
}

void change(int t,int l,int r,int x,int s)
{
	int mid=(l+r)/2;
	
	tr[t]+=s;
	
	if (l==r)
	return;
	
	if (x<=mid)
	change(t*2,l,mid,x,s);
	else
	change(t*2+1,mid+1,r,x,s);
}

int find(int t,int l,int r,int x,int y)
{
	int mid=(l+r)/2,ans=0;
	
	if (x<=l && r<=y)
	return tr[t];
	
	if (x<=mid)
	ans+=find(t*2,l,mid,x,y);
	if (mid<y)
	ans+=find(t*2+1,mid+1,r,x,y);
	
	return ans;
}

int main()
{
//	freopen("d.in","r",stdin);
	
	scanf("%d%d%d",&n,&D,&L);
	fo(i,1,n)
	{
		scanf("%d%d",&x,&y);
		
		j=x;k=y;
		x=j+k;
		y=k-j;
		
		a[++tot]={x,y,1};
		a[++tot]={x+D,y,-1};
	}
	
	sort(a+1,a+tot+1,cmp);
	
	fo(i,1,tot)
	{
		if (a[i].s==1)
		{
			ans+=sum-find(1,1,N,max(a[i].y+100001-D+1,1),min(a[i].y+100001+D-1,N));
			++sum;
		}
		
		change(1,1,N,a[i].y+100001,a[i].s);
	}
	
	printf("%lld
",ans);
}

E

用总数-没有交点的即可，删除可以看成加了一个-1的矩形

求出每个询问上下左右的矩形个数，再减去四个角上的

注意时间也算一维（不用排序），所以需要cdq

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define low(x) (x&-(x))
using namespace std;

struct type{
	int x1,y1,x2,y2,s,id;
} a[100001],b[100001];
struct Type{
	int s,id;
} c[200001];
struct type1{
	int x,s,id;
} d[100001];
struct type2{
	int x,y,s,id,Id;
} D[100001];
int tr[200001];
int ans[200001];
int n,i,j,k,l,tp,tot,Tot,sum,len;

bool Cmp(Type a,Type b) {return a.s<b.s;}
bool cmp(type2 a,type2 b) {return a.x<b.x || a.x==b.x && a.id>b.id;}

void change(int t,int s)
{
	while (t<=200000)
	{
		tr[t]+=s;
		t+=low(t);
	}
}

void clear(int t)
{
	while (t<=200000)
	{
		tr[t]=0;
		t+=low(t);
	}
}

int find(int t)
{
	int ans=0;
	
	while (t)
	{
		ans+=tr[t];
		t-=low(t);
	}
	
	return ans;
}

void work1()
{
	int i;
	
	fo(i,1,n)
	if (d[i].s)
	change(d[i].x,d[i].s);
	else
	ans[d[i].id]-=find(d[i].x-1);
	
	memset(tr,0,sizeof(tr));
}

void work2(int l,int r)
{
	int i,mid=(l+r)/2;
	
	if (l==r) return;
	
	work2(l,mid);
	work2(mid+1,r);
	
	sort(D+l,D+r+1,cmp);
	
	fo(i,l,r)
	if (D[i].Id<=mid && D[i].s)
	change(D[i].y,D[i].s);
	else
	if (D[i].Id>mid && !D[i].s)
	ans[D[i].id]+=find(D[i].y-1);
	
	fo(i,l,r)
	if (D[i].Id<=mid && D[i].s)
	clear(D[i].y);
}

int main()
{
//	freopen("e.in","r",stdin);
	
	scanf("%d",&n);
	fo(i,1,n)
	{
		scanf("%d",&tp);
		
		switch (tp)
		{
			case 1:{
				scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
				
				++sum;
				a[i].s=1;
				b[++tot]=a[i];
				break;
			}
			case 2:{
				scanf("%d",&j);
				
				--sum;
				a[i]=b[j];
				a[i].s=-1;
				break;
			}
			case 3:{
				scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
				
				a[i].id=++Tot;
				ans[Tot]=sum;
				break;
			}
		}
	}
	
//	---
	
	len=0;
	fo(i,1,n)
	{
		c[++len]={a[i].x1,i};
		c[++len]={a[i].x2,-i};
	}
	sort(c+1,c+len+1,Cmp);
	j=0;
	fo(i,1,len)
	{
		j+=i==1 || c[i].s!=c[i-1].s;
		
		if (c[i].id>0)
		a[c[i].id].x1=j;
		else
		a[-c[i].id].x2=j;
	}
	
	len=0;
	fo(i,1,n)
	{
		c[++len]={a[i].y1,i};
		c[++len]={a[i].y2,-i};
	}
	sort(c+1,c+len+1,Cmp);
	j=0;
	fo(i,1,len)
	{
		j+=i==1 || c[i].s!=c[i-1].s;
		
		if (c[i].id>0)
		a[c[i].id].y1=j;
		else
		a[-c[i].id].y2=j;
	}
	
//	---
	
	fo(i,1,n)
	if (a[i].s)
	d[i]={a[i].x2,a[i].s,0};
	else
	d[i]={a[i].x1,0,a[i].id};
	work1();
	
	fo(i,1,n)
	if (a[i].s)
	d[i]={200001-a[i].x1,a[i].s,0};
	else
	d[i]={200001-a[i].x2,0,a[i].id};
	work1();
	
	fo(i,1,n)
	if (a[i].s)
	d[i]={a[i].y2,a[i].s,0};
	else
	d[i]={a[i].y1,0,a[i].id};
	work1();
	
	fo(i,1,n)
	if (a[i].s)
	d[i]={200001-a[i].y1,a[i].s,0};
	else
	d[i]={200001-a[i].y2,0,a[i].id};
	work1();
	
//	---
	
	fo(i,1,n)
	if (a[i].s)
	D[i]={a[i].x2,a[i].y2,a[i].s,0};
	else
	D[i]={a[i].x1,a[i].y1,0,a[i].id};
	fo(i,1,n)
	D[i].Id=i;
	work2(1,n);
	
	fo(i,1,n)
	if (a[i].s)
	D[i]={200001-a[i].x1,200001-a[i].y1,a[i].s,0};
	else
	D[i]={200001-a[i].x2,200001-a[i].y2,0,a[i].id};
	fo(i,1,n)
	D[i].Id=i;
	work2(1,n);
	
	fo(i,1,n)
	if (a[i].s)
	D[i]={a[i].x2,200001-a[i].y1,a[i].s,0};
	else
	D[i]={a[i].x1,200001-a[i].y2,0,a[i].id};
	fo(i,1,n)
	D[i].Id=i;
	work2(1,n);
	
	fo(i,1,n)
	if (a[i].s)
	D[i]={200001-a[i].x1,a[i].y2,a[i].s,0};
	else
	D[i]={200001-a[i].x2,a[i].y1,0,a[i].id};
	fo(i,1,n)
	D[i].Id=i;
	work2(1,n);
	
//	---
	
	fo(i,1,Tot)
	printf("%d
",ans[i]);
}