HZOI20190823模拟31题解

题面：https://www.cnblogs.com/Juve/articles/11425141.html

math：仔细看看其实是个水题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define re register
using namespace std;
const int MAXN=5e5+5;
const int MAXM=1e6+5;
int n,k,a[MAXN],g,sum;
int gcd(int a,int b){
	return b==0?a:gcd(b,a%b);
}
signed main(){
	scanf("%lld%lld",&n,&k);
	for(re int i=1;i<=n;i++) scanf("%lld",&a[i]);
	g=k;
	for(int i=1;i<=n;i++) g=gcd(a[i],g);
	sum=k/g;
	printf("%lld\n",sum);
	for(int i=0;i*g<k;i++){
		printf("%lld ",i*g);
	}
	puts("");
	return 0;
}

biology：dp，维护4个最大值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 2005
#define re register
#define int long long
using namespace std;
int n,m,a[MAXN][MAXN],b[MAXN][MAXN],ans=0,tot=0,st;
struct node{
	int x,y,a,b;
	friend bool operator < (node p,node q){
		return p.a<q.a;
	}
}t[MAXN*MAXN];
bool vis[MAXN*MAXN];
int px1,px2,px3,px4,mx1,mx2,mx3,mx4;
signed main(){
	scanf("%lld%lld",&n,&m);
	for(re int i=1;i<=n;i++)
		for(re int j=1;j<=m;j++)
			scanf("%lld",&a[i][j]);
	for(re int i=1;i<=n;i++)
		for(re int j=1;j<=m;j++){
			scanf("%lld",&b[i][j]);
			if(a[i][j]==0) continue;
			t[++tot]=(node){i,j,a[i][j],b[i][j]};
		}
	sort(t+1,t+tot+1);
	t[0]=t[1];
	for(int i=1;i<=tot;i++) if(t[i].a!=t[i-1].a) vis[i]=1;
	for(int i=1;i<=tot;i++){
		int x=t[i].x,y=t[i].y,res=t[i].b;
		if(vis[i]){
			st=i;
			break;
		}
		px1=max(px1,res-x-y);
		px2=max(px2,res-x+y);
		px3=max(px3,res+x-y);
		px4=max(px4,res+x+y);
	}
	for(int i=st;i<=tot;i++){
		int x=t[i].x,y=t[i].y,res=0;
		if(vis[i]){
			mx1=max(mx1,px1);
			mx2=max(mx2,px2);
			mx3=max(mx3,px3);
			mx4=max(mx4,px4);
			px1=px2=px3=px4=0;
		}
		res=max(max(mx1+x+y,mx2+x-y),max(mx3-x+y,mx4-x-y))+t[i].b;
		px1=max(px1,res-x-y);
		px2=max(px2,res-x+y);
		px3=max(px3,res+x-y);
		px4=max(px4,res+x+y);
		ans=max(ans,res);
	}
	printf("%lld\n",ans);
	return 0;
}
/*
3 3
0 6 8
1 6 1
0 6 8
0 1 2
3 4 5
0 6 7
*/

english：

ans1用单调栈处理出每一个ai，它作为最大值的区间，然后维护前缀和，sum[i][j]表示前j个数第i位上有几个1

ans2用可持久化01trie做

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define int long long
using namespace std;
const int MAXN=1e5+5;
const int mod=1e9+7;
int n,a[MAXN],opt,sum[22][MAXN],ans1=0,ans2=0;
int sta[MAXN],top=0,l[MAXN],r[MAXN],tot=0;
int tr[MAXN*22][2],size[MAXN*22],root[MAXN];
void insert(int &now,int pre,int val,int i){
	now=++tot;
	size[now]=size[pre]+1;
	if(i<0) return ;
	int p=(val>>i)&1;
	tr[now][p^1]=tr[pre][p^1];
	insert(tr[now][p],tr[pre][p],val,i-1);
	size[now]=size[tr[now][0]]+size[tr[now][1]];
}
int query(int rt,int x,int y){
	int res=0;
	for(int i=20;i>=0;i--){
		int p=(x>>i)&1,q=(y>>i)&1;
		if(!q){
			(res+=size[tr[rt][p^1]])%=mod;
			rt=tr[rt][p];
		}
		else rt=tr[rt][p^1];
		if(!rt) break;
	}
	return res;
}
signed main(){
	scanf("%lld%lld",&n,&opt);
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		insert(root[i],root[i-1],a[i],20);
		int tmp=a[i];
		for(int j=0;j<=20;j++){
			sum[j][i]=sum[j][i-1];
			if((a[i]>>j)&1) sum[j][i]++;
		}
	}
	for(int i=0;i<=20;i++) sum[i][n+1]=sum[i][n],sum[i][0]=0;
	for(int i=1;i<=n;i++){
		while(top!=0&&a[sta[top]]<a[i]) top--;
		if(top==0) l[i]=1;
		else l[i]=sta[top]+1;
		sta[++top]=i;
	}
	top=0;
	for(int i=n;i>=1;i--){
		while(top!=0&&a[sta[top]]<=a[i]) top--;
		if(top==0) r[i]=n+1;
		else r[i]=sta[top];
		sta[++top]=i;
	}
	for(int i=1,res;i<=n;i++){
		res=0;
		if(l[i]==r[i]) continue;
		for(int j=0;j<=20;j++){
			(res+=(((sum[j][i]-sum[j][l[i]-1])*((r[i]-i)-(sum[j][r[i]-1]-sum[j][i-1])))<<j))%=mod;
			(res+=(((sum[j][r[i]-1]-sum[j][i-1])*((i-l[i]+1)-(sum[j][i]-sum[j][l[i]-1])))<<j))%=mod;
		}
		(ans1+=res*a[i]%mod)%=mod;
		res=0;
		if(i-l[i]<r[i]-i-1){
			for(int j=l[i];j<=i;j++)
				(res+=(query(root[r[i]-1],a[j],a[i])-query(root[i-1],a[j],a[i])+mod)%mod)%=mod;
		}else{
			for(int j=i;j<r[i];j++)
				(res+=(query(root[i],a[j],a[i])-query(root[l[i]-1],a[j],a[i])+mod)%mod)%=mod;
		}
		(ans2+=(res*a[i]%mod))%=mod;
	}
	if(opt==1) printf("%lld\n",ans1%mod);
	else if(opt==2) printf("%lld\n",ans2%mod);
	else printf("%lld\n%lld\n",ans1%mod,ans2%mod);
	return 0;
}