2018-2019 ICPC, NEERC, Southern Subregional Contest 7/13

最新推荐文章于 2019-12-16 21:07:55 发布

BPMThor(BPM136)

最新推荐文章于 2019-12-16 21:07:55 发布

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分类专栏： ACM补题

本文链接：https://blog.youkuaiyun.com/qq_20139423/article/details/86619018

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A

Solved: 01:19 (BPM136)
状态只有NM种，bfs即可

#include <bits/stdc++.h>

using namespace std;

struct node {
	int r;
	string s;
	int sum;
};

bool vis[505][5005];

int d, s;

int main() {
	cin >> d >> s;
	queue<node> Q;
	Q.push(node{0, "", 0});
	vis[0][0] = 1;
	while (Q.empty() == 0) {
		node now = Q.front();
		Q.pop();
		for (int i = 0; i < 10; ++i) {
			int nr = now.r * 10 + i;
			nr %= d;
			int ns = now.sum + i;
			if (ns > s || vis[nr][ns]) 
				continue;
			Q.push(node{nr, now.s + char(i + '0'), ns});
			vis[nr][ns] = 1;
			if (vis[0][s]) {
				cout << now.s + char(i + '0') << '\n';
				return 0;
			}
		}
	}
	puts("-1");
}
close

B

UnSolved

C

Solved: 02:09(HOOCCOOH)

#include <iostream>
#include <functional>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cinttypes>
#include <iterator>

struct Event {
    int price, cnt;
};

struct SegT {
    int s, price;
    int64_t cnt, sum;
    SegT *l, *r;

    void up() {
        cnt = l->cnt + r->cnt;
        sum = l->sum + r->sum;
    }
    void app(int dt) {
        cnt += dt;
        sum += (int64_t)price * dt;
    }
};

std::deque<SegT> buf;

SegT *make(int l, int r) {
    buf.push_back(SegT { r - l, l, 0, 0, nullptr, nullptr });
    auto c = &buf.back();
    if(r - l > 1) {
        int m = l + (r - l) / 2;
        c->l = make(l, m);
        c->r = make(m, r);
    }
    return c;
}

void add(SegT *c, int pos, int64_t dt) {
    if(c->s == 1)
        return c->app(dt);
    int m = c->s / 2;
    if(pos < m) add(c->l, pos, dt);
    else add(c->r, pos - m, dt);
    c->up();
}

int64_t select(SegT *c, int64_t cnt) {
    if(c->s == 1)
        return std::min(c->cnt, cnt) * c->price;

    if(cnt <= c->l->cnt)
        return select(c->l, cnt);
    else
        return c->l->sum + select(c->r, cnt - c->l->cnt);
}

class DualHeap {
    int k;
    SegT *rt;

public:
    DualHeap(int mxp, int k): k(k), rt(make(0, mxp + 1)) {}
    ~DualHeap() {}

    void add(Event evt) {
        ::add(rt, evt.price, evt.cnt);
    }

    int64_t get_sum() {
        return ::select(rt, k);
    }
};

int main() {
    std::ios::sync_with_stdio(false);

    int n, k, m;
    std::cin >> n >> k >> m;

    auto days = std::vector<std::vector<Event>>(n);
    int maxp = 0;
    for(int i = 0; i < m; ++i) {
        int l, r, cnt, price;
        std::cin >> l >> r >> cnt >> price;
        maxp = std::max(maxp, price);
        days[l - 1].push_back(Event { price, cnt });
        if(r < n)
            days[r].push_back(Event { price, -cnt });
    }

    int64_t ans = 0;
    int64_t cur_sum = 0, cur_cnt = 0;
    auto q = DualHeap(maxp, k);

    for(auto &day: days) {
        for(auto &evt: day)
            q.add(evt);
        ans += q.get_sum();
    }

    std::cout << ans << std::endl;

    return 0;
}

D

Solved: 00:10(zkp)
温暖签到

#include<iostream>
using namespace std;
long long a[200010];
int main()
{
	long long n,m,now=0,ans=0;
	cin>>n>>m;
	for(int i=0;i<n;i++)cin>>a[i];
	now=a[0];
	ans+=now/m;
	now%=m;
	for(int i=1;i<n;i++)
	{
		if(now!=0&&now+a[i]<m)now=0,ans++;
		else
		{
			now+=a[i];
			ans+=now/m;
			now%=m;
		}
	}
	if(now!=0)ans++;
	cout<<ans<<endl;
	return 0;
}

E

Solved: 03:58(BPM136)
事后听人说直接二分d就好了？
好像是的啊
然而我那时候觉得好像三分又觉得不能三分，更别说二分了
所以直接枚举d，然后线段树二分套线段树二分了XD
然而被卡不能输出0 0了就卡自闭了XD

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

int const N = 2e5 + 5;

ll sump[N << 2];
ll sumn[N << 2];

struct node {
	int pos;
	ll p;
};

int n, m;
ll t;

void pushup(int k) {
	sump[k] = sump[k << 1] + sump[k << 1 | 1];
	sumn[k] = sumn[k << 1] + sumn[k << 1 | 1];
}

void SegAdd(int k, int l, int r, int pos, ll p) {
	if (l == r) {
		sump[k] += p;
		sumn[k]++;
		return;
	}
	int mid = (l + r) / 2;
	if (pos <= mid)
		SegAdd(k << 1, l, mid, pos, p);
	else
		SegAdd(k << 1 | 1, mid + 1, r, pos, p);
	pushup(k);
}

pair<ll, ll> SegEventSum(int k, int l, int r, int _l, int _r) {
	if (l == _l && r == _r) 
		return make_pair(sump[k], sumn[k]);
	int mid = (l + r) >> 1;
	if (_r <= mid)
		return SegEventSum(k << 1, l, mid, _l, _r);
	if (_l > mid)
		return SegEventSum(k << 1 | 1, mid + 1, r, _l, _r);
	auto retz = SegEventSum(k << 1, l, mid, _l, mid);
	auto retr = SegEventSum(k << 1 | 1, mid + 1, r, mid + 1, _r);
	return make_pair(retz.first + retr.first, retz.second + retr.second);
}

ll SegEFByNum(int k, int l, int r, ll sumnum) {
	if (l == r) {
		if (sumn[k] > sumnum)
			return 0;
		else 
			return sump[k];
	}
	int mid = (l + r) / 2;
	if (sumn[k << 1] < sumnum)
		return sump[k << 1] + SegEFByNum(k << 1 | 1, mid + 1, r, sumnum - sumn[k << 1]);
	else
		return SegEFByNum(k << 1, l, mid, sumnum);
}

ll GetTimeOnPos(ll pos) {
	auto sumEvent = SegEventSum(1, 1, n, 1, pos);
	ll haveBreak = (sumEvent.second - 1) / m * m;
	// cerr << "To Pos " << pos << " : " << sumEvent.first << ' ' << sumEvent.second << ' ' << haveBreak << '\n';
	ll sum = SegEFByNum(1, 1, n, haveBreak);
	return sum * 2 + (sumEvent.first - sum);
}

ll SegEF(int k, int l, int r) {
	if (l == r) {
		if (GetTimeOnPos(l) <= t) 
			return sumn[k];
		else
			return 0;
	}
	int mid = (l + r) >> 1;
	ll ti = GetTimeOnPos(mid);
	if (ti < t)
		return sumn[k << 1] + SegEF(k << 1 | 1, mid + 1, r);
	else
		return SegEF(k << 1, l, mid);
}

int main() {
	ios::sync_with_stdio(0);
	int T;
	cin >> T;
	while (T--) {
		cin >> n >> m >> t;
		auto a = vector<node>(n);
		for (int i = 0; i < n; ++i) {
			cin >> a[i].p;
			a[i].pos = i + 1;
		}
		sort(a.begin(), a.end(), [] (node const& x, node const& y) {
				return x.p < y.p;
		});
		for (int i = 0; i <= n * 4; ++i)
			sump[i] = sumn[i] = 0;
		ll ans1 = 0, ans2 = 0;
		for (int i = 0; i < n; ++i) {
			auto const& ev = a[i];
			SegAdd(1, 1, n, ev.pos, ev.p);
			if (i == n - 1 || ev.p != a[i + 1].p) {
				auto su = SegEF(1, 1, n);
				// cerr << "Event : " << ev.pos << ' ' << ev.p << '\n';
				// cerr << "Finish : " << su << '\n';
				if (su > ans1) {
					ans1 = su;
					ans2 = ev.p;
				}
			}
		}
		if (ans1 == 0)
			ans2 = 1;
		cout << ans1 << ' ' << ans2 << '\n';
		// cerr << ans1 << ' ' << ans2 << '\n';
	}

	return 0;
}

F

Solved：00:26 (zkp)

#include<iostream>
#include<queue>
using namespace std;
string s[400010];
int a[400010];
priority_queue<long long>x,y,z;
int main()
{
	int n,t=0;
	long long ans=0;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>s[i]>>a[i];
		if(s[i]=="11")t++,ans+=a[i];
		if(s[i]=="10")x.push(a[i]);
		if(s[i]=="01")y.push(a[i]);
		if(s[i]=="00")z.push(a[i]);
	}
	while(!x.empty()&&!y.empty())
	{
		ans+=x.top()+y.top();
		x.pop();y.pop();
	}
	while(t--)
	{
		if(x.empty())x.push(0);
		if(y.empty())y.push(0);
		if(z.empty())z.push(0);
		if(x.top()>=y.top()&&x.top()>=z.top())
		{
			ans+=x.top();
			x.pop();
		}
		else if(y.top()>=x.top()&&y.top()>=z.top())
		{
			ans+=y.top();
			y.pop();
		}
		else
		{
			ans+=z.top();
			z.pop();
		}
	}
	cout<<ans<<endl;
	return 0;
}

G

Unsolved

H

Solved: 01:24(zkp)
dirt：我的锅，用了奇怪的写法写完后无限RE on test1，但是本地测试没问题，后果断让zkp重写了

#include <bits/stdc++.h>
using namespace std;
int a[400]={};
map<unsigned long long,pair<int,string> >mp;
using namespace std;
int main() {
	ios::sync_with_stdio(0);
	unsigned long long t=1;
	int n,m;
	for(int i='a';i<='z';i++,t++)a[i]=t;
	for(int i='0';i<='9';i++,t++)a[i]=t;
	for(int i='.';i<='.';i++,t++)a[i]=t;
	string s;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>s;
		for(int j=0;j<s.size();j++)
		{
			unsigned long long now=0;
			for(int k=j;k<s.size();k++)
			{
				now=now*t+a[s[k]];
				if(mp[now].second!=s)mp[now].first++;
				mp[now].second=s;
			}
		}
	} 
	cin>>m;
	while(m--)
	{
		cin>>s;
		unsigned long long now=0;
		for(int i=0;i<s.size();i++)now=now*t+a[s[i]];
		if(mp[now].first!=0)cout<<mp[now].first<<" "<<mp[now].second<<endl;
		else cout<<"0 -"<<endl;
	}
	return 0;
}

I

Unsolved

J

Unsolved(-5)
只有一种字母会同时在左右两边
枚举这个字母，然后背包一下

K

Solved: 00:07(BPM136)

#include <bits/stdc++.h>

using namespace std;

int main() {
	int n, m;
	cin >> n >> m;
	auto a = vector<int>();
	long long sum = 0;
	for (int i = 0; i < n; ++i) {
		int x;
		cin >> x;
		a.push_back(x);
		sum += x;
	}
	if (sum % m) 
		puts("No");
	else {
		int fen = sum / m;
		sum = 0;
		bool can = 1;
		auto ans = vector<int>();
		int num = 0;
		for (int i = 0; i < n; ++i) {
			sum += a[i];
			++num;
			if (sum > fen) {
				can = 0;
				break;
			}
			if (sum == fen) {
				ans.push_back(num);
				sum = 0;
				num = 0;
				continue;
			}
		}
		if (can) {
			puts("Yes");
			for (auto& v : ans)
				cout << v << ' ';
			cout << '\n';
		} else
			puts("No");
	}
	return 0;
}

L

Unsolved

M

Unsolved