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2020-2021 “Orz Panda” Cup Programming Contest

作者:互联网

2020-2021 “Orz Panda” Cup Programming Contest

A. Accordion Artist And Orz Pandas

题意

给定\(n\)个相邻的矩形,矩形的宽和高给定,问最大的子矩形面积

分析

单调栈板题,维护出每个矩形左边第一个比他低的点和右边第一个比它低的点,两点距离就是宽,高就是当前矩形的高,维护宽X高的最大值即可

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll rd(){
	ll x;
	scanf("%lld",&x);
	return x;
}


int main(){
	int n = rd();
	vector<ll> h(n + 1),w(n + 1);
	vector<ll> sum(n + 1);
	vector<int> lpos(n + 1),rpos(n + 1);
	for(int i = 1;i <= n;i++){
		h[i] = rd();
		w[i] = rd();
		sum[i] = sum[i - 1] + w[i];
	}
	stack<int> st;
	for(int i = 1;i <= n;i++){
		while(!st.empty() && h[st.top()] > h[i]) {
			rpos[st.top()] = i;
			st.pop();
		}
		st.push(i);
	}
	while(!st.empty()) {
		rpos[st.top()] = n + 1;
		st.pop();
	}
	for(int i = n;i >= 1;i--){
		while(!st.empty() && h[st.top()] > h[i]) {
			lpos[st.top()] = i;
			st.pop();
		}
		st.push(i);
	}
	while(!st.empty()){
		lpos[st.top()] = 0;
		st.pop();
	}
	ll ans = 0;
	for(int i = 1;i <= n;i++){
		ll W = sum[rpos[i] - 1] - sum[lpos[i]];
		ans = max(ans,W * h[i]);
	}
	cout << ans;
}

C. Closestools of Orz Pandas

题意

模拟题

给定\(n\)个房间,定义i,j号房间距离\(|i - j|\)

操作1:选择一个空房间进入,要求该空房间得到的和其他非空房间生成的距离排序后字典序最小

操作2:将第x次操作的房间变为空房间

分析

贪心策略:

把空房间看成线段,每次找到最长的线段,取最长线段的中点。

通常维护线段可以用set实现。再用另一个线段维护非空房间,可以快速得到左右点位置。

代码

#include<bits/stdc++.h>
#define re register
using namespace std;
typedef long long ll;


int rd(){
	int x = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x;
}

struct Seg{
	int l,r; //[l,r]
	Seg(int _l = 0,int _r = 0){
		l = _l;
		r = _r;
	}
	int d() const {
		return (r - l) / 2 + 1;
	}
	bool operator < (const Seg& b) const{
		int d1 = d();
		int d2 = b.d();
		return (d1 > d2) || (d1 == d2 && l < b.l);
	}
};


int main(){
	int n,m;
	while(~scanf("%d%d",&n,&m)){
		set<Seg> s;
		s.insert(Seg(1,n));
		set<int> st;
		vector<int> v(m + 1);
		for(re int i = 1;i <= m;i++){
			int op = rd();
			if(op == 1) {
				auto best = s.begin();
				int bestd = best -> d();
				if(!st.empty() && *st.begin() > bestd) {
					best = s.lower_bound(Seg(1,*st.begin() - 1));
					bestd = (*st.begin() - 1) / 2;
				}
				if(!st.empty() && n - *st.rbegin() > bestd) {
					best = s.lower_bound(Seg(*st.rbegin() + 1,n));
					bestd = (n - *st.rbegin()) / 2;
				}
				if(st.empty()) 
					best = s.begin();
				int pos = best -> l + bestd - 1;
				if(best -> l == 1)
					pos = 1;
				else if(best -> r == n)
					pos = n;
				printf("%d\n",pos);
				v[i] = pos;
				st.insert(pos);
				Seg L = Seg(best -> l,pos - 1);
				Seg R = Seg(pos + 1,best -> r);
				s.erase(best);
				if(L.l <= L.r) 
					s.insert(L);
				if(R.l <= R.r)
					s.insert(R);
			}
			else {
				int x = rd();
				int pos = v[x];
				auto it = st.lower_bound(pos);
				auto itt = it;
				int L = (it != st.begin() ? *(--it) + 1 : 1);
				itt++;
				int R = (itt != st.end() ? *itt - 1 : n);
				if(L <= pos - 1) 
					s.erase(Seg(L,pos - 1));
				if(pos + 1 <= R)
					s.erase(Seg(pos + 1,R));
				s.insert(Seg(L,R));
				st.erase(pos);
			}
		}
	}
}

D. Data Structure Master and Orz Pandas

题意

大概意思就是求期望(如有不对的地方请指正)

对一颗树中的某点,找到根到这个点的路径把这条路径上的边都变成实边(LCT中,某个点的儿子中只能连一条实边),类似LCT的access操作

求稳态下拉一个点,边从虚边变为实边的期望次数

分析

考虑答案组成,枚举每个点,那么这个点到路径上所有点被修改的期望次数的和就是答案(期望的可加性)

\[ans = \frac{1}{n}\sum_u \sum_{v是u到root的路径上的点} E(v和fa的边不是实边) \]

考虑\(E\),显然只和\(v\)和\(fa_v\)有关,简单推导(直觉)就可以得到\(E = 1 - \frac{siz(v)}{siz(fa_v) - 1}\)

于是变换求和顺序即可得

\[ans = \frac{1}{n} \sum_u siz(u) (1 - \frac{siz(u)}{siz(fa_u) - 1}) \]

dfs即可

代码

#include<bits/stdc++.h>
#define re register
using namespace std;
typedef long long ll;


int rd(){
	int x = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x;
}

const int maxn = 1e5 + 5;
const int MOD = 998244353;

ll ksm(ll a,ll b = MOD - 2,ll m = MOD){
	ll base = a;
	ll ans = 1;
	while(b){
		if(b & 1) ans *= base,ans %= MOD;
		base *= base;
		base %= MOD;
		b >>= 1;
	}
	return ans;
}

int Fa[maxn];
int siz[maxn];
vector<int> e[maxn];
int ans;

void dfs(int u,int fa){
	Fa[u] = fa;
	siz[u] = 1;
	for(auto v:e[u]) {
		dfs(v,u);
		siz[u] += siz[v];		
	}
}


int main(){
	int n = rd();
	for(int i = 2;i <= n;i++){
		int x = rd();
		e[x].push_back(i);		
	}
	dfs(1,0);
	for(int u = 2;u <= n;u++){
		 ans += (ll)siz[u] * (1 - (ll)siz[u] * ksm(siz[Fa[u]] - 1 + MOD) % MOD + MOD) % MOD,ans %= MOD;
	}
	ans = (ll) ans * ksm(n) % MOD;
	cout << ans;
}

E. Encryption of Orz Pandas

题意

经过转化可以把问题变为:

给定长度\(n\)的01序列,对序列做k次前缀和,求k次前缀和后的序列

\(n \leq 1e5\)

分析

做k次前缀和可以用生成函数看成乘上\((1 + x+ x^2...+x^n)^k\)

即若有序列\((a_0,a_1...a_n)\),那么k次前缀和后的\(a_i = [x^i](a_0 + a_1 ...+a_n) \times (1+x+x^2...x^n)^k\)

做法可以用题解写的泰勒展开后转化为组合数再FFT,复杂度\(O(nlogn)\)

不过我看不懂,比较朴素的想法是对\(FFT\)做快速幂复杂度\(O(nlognlogk)\) 考虑到时限5s可以冲一冲

我觉得FFT调试起来有点麻烦,事实上当时没调出来,赛后发现用NTT即可,01序列做1次不会爆模数

代码

#include<bits/stdc++.h>
#define re register
using namespace std;
typedef long long ll;


ll rd(){
	ll x = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x;
}

const int maxn = 2e5 + 5;

class NTTClass{
public:
    static const int MAXL=22;
    static const int MAXN=1<<MAXL;
    static const int root=3;
    static const int MOD=998244353;
    int rev[MAXN];
 
    int fast_pow(int a,int b){
        int ans=1;
        while(b){
            if(b&1)ans=1ll*ans*a%MOD;
            a=1ll*a*a%MOD;
            b>>=1;
        }
        return ans;
    }
 
    void transform(int n,int *t,int typ){
        for(int i=0;i<n;i++)
            if(i<rev[i])swap(t[i],t[rev[i]]);
        for(int step=1;step<n;step<<=1){
            int gn=fast_pow(root,(MOD-1)/(step<<1));
            for(int i=0;i<n;i+=(step<<1)){
                int g=1;
                for(int j=0;j<step;j++,g=1ll*g*gn%MOD){
                    int x=t[i+j],y=1ll*g*t[i+j+step]%MOD;
                    t[i+j]=(x+y)%MOD;
                    t[i+j+step]=(x-y+MOD)%MOD;
                }
            }
        }
        if(typ==1)return;
        for(int i=1;i<n/2;i++)swap(t[i],t[n-i]);
        int inv=fast_pow(n,MOD-2);
        for(int i=0;i<n;i++)t[i]=1ll*t[i]*inv%MOD;
    }
 
    void ntt(int p,int *A,int *B,int *C){
        transform(p,A,1);
        transform(p,B,1);
        for(int i=0;i<p;i++)C[i]=1ll*A[i]*B[i]%MOD;
        transform(p,C,-1);
    }
 
    void mul(int *A,int *B,int *C,int n,int m) {
        int p=1,l=0;
        while(p<=n+m)p<<=1,l++;
        //printf("n = %d, m = %d\n",n,m);
        for (int i=n+1;i<p;i++) A[i] = 0;
        for (int i=m+1;i<p;i++) B[i] = 0;
        //for (int i=0;i<p;i++) printf("%d ",A[i]);puts("");
        //for (int i=0;i<p;i++) printf("%d ",B[i]);puts("");
        for(int i=0;i<p;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
        ntt(p,A,B,C);
        //puts("C:");for (int i=0;i<p;i++) printf("%d ",C[i]);puts("");
    }
}NTT;

int _[maxn << 1],f[maxn << 1];
int ans[maxn << 1],p[maxn << 1],ttt[maxn << 1];
int n,m,mm;

inline void ksm(int *a,ll b){
	ans[0] = 1;
	for(re int i = 1;i < n;i++)
		ans[i] = 0;
	while(b){
		if(b & 1){
			for(re int i = 0;i < n;i++)
				p[i] = a[i];
			NTT.mul(ans,a,_,n,n);
			for(re int i = 0;i < n;i++)
				a[i] = p[i],ans[i] = (_[i] & 1 ? 1 : 0);
		}
		for(re int i = 0;i < n;i++)
			p[i] = a[i];
		NTT.mul(a,p,_,n,n);
		for(re int i = 0;i < n;i++)
			a[i] = (_[i] & 1 ? 1 : 0);
		//for(re int i = 0;i <= mm;i++) {
		//	int tmp = (int)(ans[i].x / n + 0.49);
		//	if(tmp & 1) ans[i].x = 1;
		//	else ans[i].x = 0;
		//}
	   	b >>= 1;	
	}
}

int main(){
	n = rd();
	mm = n;
	m = n;
	ll k = rd();
	int cnt = 0;
	int M = mm;
	while(M) M /= 2,cnt++;
	if((1ll << cnt) != mm) cnt++;
   	k = (k - 1) % (1 << cnt);
	k++;
	vector<ll> v(mm);
	vector<ll> res(mm);
	for(int i = 0;i < mm;i++)
		v[i] = rd();
	for(int i = 0;i < n;i++)
		f[i] = 1;
	ksm(f,k);
	//for(int i = 0;i <= mm;i++)
	//	printf("%d ",ans[i]);
	
	for(int i = 0;i <= 17;i++){
		for(re int j = 0;j < mm;j++){
			int tttt = ((v[j] >> i) & 1);
			ttt[j] = ans[j];
			f[j] = tttt;
		}
		for(int j = mm;j < n;j++){
			ttt[j] = f[j] = 0;
		}
		/*
		if(i == 2) for(int j = 0;j < n;j++){
			cout << ttt[j].x << ' ' << f[j].x << '\n';
		}*/
		NTT.mul(f,ttt,_,n,n);
		for(re int j = 0;j < mm;j++){
			int ok = _[j] & 1 ? 1 : 0;
			//if(ok) cout << "i = " << i << "j = " << j << '\n';
			if(ok) res[j] |= (1ll << i);
		}
	}
	for(int i = 0;i < mm;i++)
		if(i) putchar(' '),cout << res[i];
		else cout << res[i];
	
}

H. Hamming Code and Orz Pandas

模拟题,因为我没有参与此题,所以无法提供解释

代码

#include<bits/stdc++.h>
#define ll long long
using namespace std;



ll rd(){
	ll x = 0;
	int f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * f;
}

int k;
string d;
vector<int>a;
vector<int>have[20];
void init()
{
    int num=1;
    a.push_back(0);
    for(int i=1;i<=17;i++){
        a.push_back(num);
        num=num*2;
    }
    int p=1;
    for(int i=3;i<=1024*64*2;i++){
        p=i;
        for(int j=16;j>=1;j--){
            if(i==a[j]){
                p=0;
                break;
            }
        }
        if(p==0)continue;
        int hh=0;
        while(p){
            for(int j=17;j>=1;j--){
                hh=0;
                while(p>=a[j]){
                    p-=a[j];
                    hh=1;
                }
                if(hh)have[j].push_back(i);
                if(p==0)break;
            }
        }
    }
}
int main()
{
    init();

    while(~scanf("%d",&k)){
        cin>>d;
        int num=0;
        for(int i=0;i<d.size();i++){
            num=num^d[i];
        }
        if(num==0){
            int wa=0;
            for(int i=1;i<=k;i++){
                int pos=0;
                //cout<<i<<":"<<endl;
                for(auto x:have[i]){

                    if(x>=a[k+1])break;
                    //cout<<x<<endl;
                    pos=pos^d[x];
                }
                if(pos!=d[a[i]]){
                    wa=1;
                    break;
                }
            }
            if(wa){
                printf("broken\n");
            }else{
                printf("good\n");
            }
        }else{
            vector<int>ans;
             for(int i=1;i<=k;i++){
                int pos=0;
                for(auto x:have[i]){
                    if(x>=a[k+1])break;
                    pos=pos^d[x];
                }
                if(pos!=d[a[i]]){
                    ans.push_back(a[i]);
                }
            }
            //cout<<ans.size()<<endl;
            if(ans.size()==1){
                printf("d(%d) is changed\n",ans[0]);
            }else{
                int sum=0;
                for(auto x:ans){
                    sum+=x;
                }
                printf("d(%d) is changed\n",sum);
            }
        }
    }
}
/*
4
1011101110111011
4
1011101110111010
4
1011101110111110


*/

I. Irregular Shape of Orz Pandas

题意

顺序给出点,求多边形面积

\[3 \leq n \leq 1e3\\ -1e9 \leq x,y \leq 1e9 \]

分析

一开始用double算,发现爆了,因为给的是整点,所以直接用__int128,事实上longlong即可,最后除以二可以人为加小数 这题花了很长时间,实在是不应该

代码

#include<bits/stdc++.h>
using namespace std;
typedef __int128 ll;

ll rd(){
	ll x = 0;
	int f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		if(ch == '-') f = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * f;
}

void Put(ll x){
    if(x == 0) return;
	if(x < 10) {
		putchar(x + '0');
		return;
	}
    Put(x / 10);
	putchar(x % 10 + '0');
}

struct Point{
	ll x;
	ll y;
	Point(){}
	Point(ll _x,ll _y):x(_x),y(_y){}
};
ll operator ^ (const Point &a,const Point &b){
	return a.x * b.y - a.y * b.x;
}

Point operator - (const Point &a,const Point &b){
	return Point(a.x - b.x,a.y - b.y);
}

struct Pol{
	int n;
	Point p[1005];
	void input(int _n){
		n  = _n;
		for(int i = 0;i < n;i++){
			p[i].x = rd();
			p[i].y = rd();
		}
	}
	ll getarea(){
		ll sum = 0;
		for(int i = 0;i < n;i++){
			sum += ((p[i]) ^ (p[(i + 1) % n]));
		}
		return sum; 
	}
};

int main(){
	int n;
	while(~scanf("%d",&n)){
		Pol poly;
		poly.input(n);
        ll tmp = poly.getarea();
        if(tmp < 0) tmp *= -1;
        if(tmp % 2 == 1) {
            Put(tmp / 2);
            putchar('.');
            putchar('5');
            putchar('0');
        }
        else {
            Put(tmp / 2);            
            putchar('.');
            putchar('0');
            putchar('0');
        }
        puts("");
    }
}

标签:ch,Orz,Contest,int,ll,Cup,st,rd,while
来源: https://www.cnblogs.com/hznumqf/p/14969493.html