20180720 考试记录

T1 Black Rock Shooter

Solution

GSS3原题qwq

Code

//By Menteur_Hxy
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define int long long
#define ls (cur<<1)
#define rs (cur<<1|1)
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

int read() {
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int INF=0x3f3f3f3f;
const int N=500010;
int n,m,asu,amx,als,ars,ans;
int sum[N<<2],lsu[N<<2],rsu[N<<2],mx[N<<2];

void pushup(int cur) {
	sum[cur]=sum[ls]+sum[rs];
	lsu[cur]=max(lsu[ls],sum[ls]+lsu[rs]);
	rsu[cur]=max(rsu[rs],sum[rs]+rsu[ls]);
	mx[cur]=max(max(mx[ls],mx[rs]),max(max(lsu[ls],rsu[rs]),rsu[ls]+lsu[rs]));
} 

void build(int cur,int l,int r) {
	if(l==r) {sum[cur]=lsu[cur]=rsu[cur]=mx[cur]=read();return ;}
	int mid=(l+r)>>1;
	build(ls,l,mid);
	build(rs,mid+1,r);
	pushup(cur);
}

void merge(int cur) {
	if(asu==-INF) asu=sum[cur],amx=mx[cur],als=lsu[cur],ars=rsu[cur];
	else {
		int a,b,c,d;
		a=asu+sum[cur];
		b=max(als,asu+lsu[cur]);
		c=max(rsu[cur],ars+sum[cur]);
		d=max(max(amx,mx[cur]),max(max(als,rsu[cur]),ars+lsu[cur]));
		asu=a,als=b,ars=c,amx=d;
	}
	ans=max(max(als,ars),max(ans,amx));
} 

void query(int cur,int l,int r,int L,int R) {
	if(L<=l&&r<=R) {merge(cur);return ;}
	int mid=(l+r)>>1;
	if(L<=mid) query(ls,l,mid,L,R);
	if(R>mid) query(rs,mid+1,r,L,R);
}

void update(int cur,int l,int r,int x,int k) {
	if(l==x&&r==x) {sum[cur]=k,lsu[cur]=k,rsu[cur]=k,mx[cur]=k;return ;}
	int mid=(l+r)>>1;
	if(x<=mid) update(ls,l,mid,x,k);
	else if(x>mid) update(rs,mid+1,r,x,k);
	pushup(cur);
}

signed main() {
	freopen("BRS.in","r",stdin);
	freopen("BRS.out","w",stdout);
	n=read(),m=read();
	build(1,1,n);
	F(i,1,m) {
		ans=asu=amx=als=ars=-INF;
		int opt=read(),x=read(),y=read();
		if(opt==1) query(1,1,n,x,y),printf("%lld
",ans);
		else update(1,1,n,x,y);
	}
	return 0;
}

T2 Fy's dota2

Solution

类似斐波那契的构造矩阵就行
(注意不要赋错初值qwq)

Code

//By Menteur_Hxy
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define int long long
#define ls (cur<<1)
#define rs (cur<<1|1)
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define R(i,a,b) for(register int i=(b);i>=(a);i--)
using namespace std;

int read() {
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int MOD=7777777;
struct Matrix{
	int da[11][11];
	void clear() {F(i,1,10) F(j,1,10) da[i][j]=0;}
	Matrix operator * (const Matrix oth) {
		Matrix ret;ret.clear();
		F(i,1,10) F(j,1,10) F(k,1,10) 
			ret.da[i][j]+=da[i][k]*oth.da[k][j]%MOD,ret.da[i][j]%=MOD;
		return ret;
	} 
	void print() {
		F(i,1,10) {
			F(j,1,10) cout<<da[i][j]<<" ";
			cout<<endl;
		}cout<<endl;
	}
};

Matrix Qpow(Matrix ret,int d) {
	Matrix res; res.clear();
	F(i,1,10) res.da[i][i]=1;
	while(d) {
		if(d&1) res=res*ret;
		ret=ret*ret; d>>=1;
	}
	return res;
}

signed main() {
//	freopen("fyfy.in","r",stdin);
//	freopen("fyfy.out","w",stdout);
	int k=read(),n=read();
	Matrix nxt;nxt.clear();
	F(i,1,k) nxt.da[i][1]=1;
	F(i,1,k) nxt.da[i][i+1]=1;
	Matrix ans=Qpow(nxt,n);
	printf("%lld",ans.da[1][1]%MOD);
	return 0;
}

T3 Olddriver’s books

Solution

好像叫做扫描线什么的，这道题比较水n^{3就能过(我写的n}2)
(注意不要少想一种情况qwq)

Code

//By Menteur_Hxy
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#define int long long
#define ls (cur<<1)
#define rs (cur<<1|1)
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define R(i,a,b) for(register int i=(b);i>=(a);i--)
using namespace std;

int read() {
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f; c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int INF=0x3f3f3f3f;
const int N=1100;
int n,cnt;
int hi[N],lo[N];
int vis[N];
struct xid{int da,id;}idx[N<<1];
struct yid{int da,fl,id;}idy[N<<1];

bool cmp1(xid a,xid b) {return a.da<b.da;} 
bool cmp2(yid a,yid b) {return a.da<b.da;}

signed main() {
	freopen("olddriver.in","r",stdin);
	freopen("olddriver.out","w",stdout);
	n=read();
	F(i,1,n) {
		idx[++cnt].da=read(),idx[cnt].id=i;
		idy[cnt].da=read(),idy[cnt].fl=1,idy[cnt].id=i;
		idx[++cnt].da=read(),idx[cnt].id=i;
		idy[cnt].da=read(),idy[cnt].fl=-1,idy[cnt].id=i;
	}
	sort(idx+1,idx+1+cnt,cmp1);
	sort(idy+1,idy+1+cnt,cmp2);
	int ans=0;
	F(i,1,cnt) {
		int sum=0,l=0,tmp=0;
		F(j,1,cnt) if(vis[idy[j].id]) {
			if(!tmp) l=idy[j].da;
			tmp+=idy[j].fl;
			if(!tmp) sum+=idy[j].da-l;
		}
		ans+=sum*(idx[i].da-idx[i-1].da);
		vis[idx[i].id]^=1ll;
	}
	printf("%lld",ans);
	return 0;
}