NOIP 模拟赛 4

NOIP 模拟赛总结

NOIP 模拟赛 4

4.5小时T3写不完！？码力还是不够啊

T1 括号问号

简单DP题，写着写着就出来了。

点击查看代码

#include<bits/stdc++.h>
#define int long long
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
#define Blue_Archive return 0
using namespace std;
constexpr int N = 5e3 + 7;
constexpr int mod = 998244353;

int n;
int ans;
int dp[N][N];

string s;

inline int read() // 不读负数的普通快读
{
	int k = 0,f = 1;
	int c = getchar_unlocked();
	while(c < '0' || c > '9') c = getchar_unlocked();
	while(c >= '0' && c <= '9') k = (k << 3) + (k << 1) + (c ^ 48),c = getchar_unlocked();
	return k * f;
}

inline void write(int x) // 普通快写
{
	if(x < 0) putchar_unlocked('-'),x = -x;
	if(x > 9) write(x / 10);
	putchar_unlocked(x % 10 + '0');
}

signed main()
{
	freopen("bracket.in","r",stdin);freopen("bracket.out","w",stdout);
	// freopen("data.in","r",stdin);freopen("data.out","w",stdout);
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> n;
	cin >> s;
	s = ' ' + s;
	dp[0][0] = 1;
	for(int i = 1;i <= n;i ++)
	{
		for(int j = 0;j <= i;j ++)
		{
			dp[i][j] = dp[i - 1][j];
			if(s[i] == '(' && j > 0) (dp[i][j] += dp[i - 1][j - 1]) %= mod;
			if(s[i] == ')') (dp[i][j] += dp[i - 1][j + 1]) %= mod;
			if(s[i] == '?')
			{
				(dp[i][j] += dp[i - 1][j + 1]) %= mod;
				if(j > 0) (dp[i][j] += dp[i - 1][j - 1]) %= mod;
			} 
		}
	}
	cout << dp[n][0] << '\n';
	Blue_Archive;
}

T2 狗卡

神秘贪心。

维护一个凸包。

点击查看代码

#include<bits/stdc++.h>
#define int long long 
#define Blue_Archive return 0
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
using namespace std;
constexpr int N = 1.2e6 + 7;
constexpr int M = 6e5 + 7;

int n;
int m;
int sm;
int ans;
int top;
int tpp;
int tim;
int k[M];
int stk[N];
int val[N];
int sumk[M];

vector<int> sum[M],a[M];

struct miku
{
	int id,l,r,val,len;
	miku(int Id = 0,int L = 0,int R = 0)
	{
		if(!Id)	return;
		id = Id,l = L,r = R;
		val = sum[id][R] - sum[id][L - 1],len = R - L + 1;
	}
	friend bool operator<(miku x,miku y){return x.val * y.len < x.len * y.val;}
}sta[N];

inline int read()
{
	int x = 0,f = 1;
	int c = getchar_unlocked();
	while(c < '0' || c > '9') f = (c == '-' ? -1 : f),c = getchar_unlocked();
	while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48),c = getchar_unlocked();
	return x * f;
}

inline void write(int x)
{
	if(x < 0) putchar_unlocked('-'),x = -x;
	if(x > 9) write(x / 10);
	putchar_unlocked(x % 10 + '0');
}

signed main()
{
	// freopen("data.in","r",stdin);freopen("data.out","w",stdout);
	freopen("dog.in","r",stdin),freopen("dog.out","w",stdout);
	n = read();
	m = read();
	for(int i = 1;i <= n;i ++)
	{
		k[i] = read();
		sumk[i] = sumk[i - 1] + k[i];
		a[i].emplace_back(0);
		sum[i].emplace_back(0);
		for(int j = 0,x;j < k[i];j ++)
		{
			x = read();
			a[i].emplace_back(x);
			sum[i].emplace_back(x + sum[i][j]);
		}
		top = 0;
		stk[++ top] = 1;
		val[1] = sum[i][1];
		for(int j = 2;j <= k[i];j ++)
		{
			while(top && (sum[i][j] - val[top]) * (stk[top] - stk[top - 1]) <= (val[top] - val[top - 1]) * (j - stk[top])) top --;
			stk[++ top] = j;
			val[top] = sum[i][j];
		}
		for(int j = 1;j <= top;j ++) sta[++ tpp] = miku(i,stk[j - 1] + 1,stk[j]);
		sm += sum[i][k[i]];
	}
	sort(sta + 1,sta + tpp + 1);
	for(int i = 1;i <= tpp;i ++)
	{
		for(int j = sta[i].l;j <= sta[i].r;j ++)
		{
			ans += (tim ++) * a[sta[i].id][j];
		}
	}
	write(ans + tim * (m - sm));ent;
	Blue_Archive;
}

T3：均衡区间

场上没写完，呜呜呜

考虑扫描线，枚举一个端点，求以其为一端的贡献。

发现贡献可以拆分成两部分。

一部分是用单调栈维护比它大或小的数的个数。

发现这样会多算，所以第二部分要斥一下，斥掉以现在枚举点为最近的最值的区间。

神秘 thick：两个单调栈下标取 \(\max\) ，之后的所有值均合法，就是最后的答案。

点击查看代码

#include<bits/stdc++.h>
#define int long long 
#define con putchar_unlocked(' ')
#define ent putchar_unlocked('\n')
#define Blue_Archive return 0
using namespace std;
constexpr int N = 1e6 + 7;
constexpr int LG = 21;
constexpr int INF = 1e9;
constexpr int mod = 998244353;

int n;
int tpy;
int top1;
int top2;
int a[N];
int ans1[N];
int ans2[N];
int stk1[N];
int stk2[N];

inline int read() // 不读负数的普通快读
{
	int k = 0,f = 1;
	int c = getchar_unlocked();
	while(c < '0' || c > '9') c = getchar_unlocked();
	while(c >= '0' && c <= '9') k = (k << 3) + (k << 1) + (c ^ 48),c = getchar_unlocked();
	return k * f;
}

inline void write(int x) // 普通快写
{
	if(x < 0) putchar_unlocked('-'),x = -x;
	if(x > 9) write(x / 10);
	putchar_unlocked(x % 10 + '0');
}

signed main()
{
	freopen("interval.in","r",stdin);freopen("interval.out","w",stdout);
	// freopen("data.in","r",stdin);freopen("data.out","w",stdout);
	n = read();
	tpy = read();
	for(int i = 1;i <= n;i ++) a[i] = read();
	if(tpy == 2)
	{
		for(int i = 1;i <= n;i ++) write(0),con;ent;
		for(int i = 1;i <= n;i ++) write(0),con;ent;
		exit(0);
	}
	top1 = top2 = 0;
	for(int i = 1,l,r,ans;i <= n;i ++)
	{
		l = r = i;
		while(a[l] == a[r + 1]) r ++;
		while(top1 && a[stk1[top1]] > a[i]) top1 --;
		while(top2 && a[stk2[top2]] < a[i]) top2 --;
		ans = 0;
		if(top1 && top2)
		{
			int op = min(stk1[top1],stk2[top2]);
			int cnt1 = upper_bound(stk1 + 1,stk1 + top1 + 1,op) - stk1 - 1;
			int cnt2 = upper_bound(stk2 + 1,stk2 + top2 + 1,op) - stk2 - 1;
			ans = op - cnt1 - cnt2;
		}
		for(int j = l;j <= r;j ++)
		{
			ans2[j] = ans;
			stk1[++ top1] = j;
			stk2[++ top2] = j;
		}
		i = r;
	}
	reverse(a + 1,a + 1 + n);
	top1 = top2 = 0;
	for(int i = 1,l,r,ans;i <= n;i ++)
	{
		l = r = i;
		while(a[l] == a[r + 1]) r ++;
		while(top1 && a[stk1[top1]] > a[i]) top1 --;
		while(top2 && a[stk2[top2]] < a[i]) top2 --;
		ans = 0;
		if(top1 && top2)
		{
			int op = min(stk1[top1],stk2[top2]);
			int cnt1 = upper_bound(stk1 + 1,stk1 + top1 + 1,op) - stk1 - 1;
			int cnt2 = upper_bound(stk2 + 1,stk2 + top2 + 1,op) - stk2 - 1;
			ans = op - cnt1 - cnt2;
		}
		for(int j = l;j <= r;j ++)
		{
			ans1[j] = ans;
			stk1[++ top1] = j;
			stk2[++ top2] = j;
		}
		i = r;
	}
	reverse(ans1 + 1,ans1 + 1 + n);
	for(int i = 1;i <= n;i ++) write(ans1[i]),con;ent;
	for(int i = 1;i <= n;i ++) write(ans2[i]),con;ent;
	Blue_Archive;
}

T4：喵了个喵了个喵

meow~