[考试总结]noip模拟28

时间:2021-07-31
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终于改掉了一看题目就想暴力的毛病

这次题目背景真的很有感觉。。

所以我无情地将其扒下来了。。。

“ 这个是为了发动奇迹所必要的准备阶段,是比较特殊的符卡。

明明就是准备的仪式却做成弹幕,真是聪明啊。

” ——《魔理沙的魔法书》


“ 虽然不清楚是不是那两人的力量

在那个风暴肆虐的夜晚,的确有一瞬

真的在那一瞬间,在云破天开的时候

透过空隙中看到的璀璨星空,不知为何倒映眼中不能忘怀 ” ——《奇迹召唤星辰》


“ 注视着属于我 独一无二的命运

席卷劲风 摧破结界

不再束缚于常识之海的我

撕裂大海 去开辟真正的道路吧

” ——《无尽的风之轨迹》


不水了

\(T1\) 其实有个很显然的贪心,然后这个题也就没什么了,然后复杂度虽然我感觉不太对,然而考场上下发的打样例秒出了。。

所以感觉很稳。。

然而 \(T2\) 给我来了当头一棒。。。

暴力都 \(TM\) 打不对,然后经过我 \(2h\) 的努力,终于调出来了暴力。

然后只有一点点时间的时候开的 \(T3\),然后发现题目很难懂,其实是因为太中二了。。

但是最后还是过了大样例。。。

遗忘之祭仪

这个题目就是记录下来附节上的所有x的位置。

然后对给出的大图进行遍历,之见到一个x就要去填满,如果发现不合法,直接puts("No");就行了。



%: pragma GCC optimize("O9")
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define debug cout<<"debug"<<endl
FILE *xinnb; int ak;
void openfile() {xinnb = freopen("t.txt","r",stdin);}
static const int maxn = 1e3+10,inf = 1e9+7; 
namespace xin
{
	char s[maxn];
	int T,n,m,hang,lie;
	int a[maxn][maxn];
	class xin_data
	{
		public:
			int x,y;
			xin_data(){}
			xin_data(int x,int y):x(x),y(y){}
	}d[maxn*maxn];int zhi = 0;
	inline void outa()
	{
		try(i,1,n)
		{
			try(j,1,m) cout<<a[i][j]<<' ';
			cout<<endl;
		}
		cout<<endl;
	}
	inline bool check(int x,int y)
	{
		try(i,1,zhi)
		{
			register int dx = x + d[i].x - 1,dy = y + d[i].y - 1;
//			cout<<"dx = "<<dx<<" dy = "<<dy<<endl;
//			outa();
			if(dx > n or dy > m or !a[dx][dy]) {return false;}
			a[dx][dy] = 0;
		}
		return true;
	}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		ak = scanf("%d",&T);	
		while(T--)
		{
			zhi = 0;
			ak = scanf("%d%d%d%d",&n,&m,&hang,&lie);
			try(i,1,n) 
			{
				ak = scanf("%s",s+1);
				try(j,1,m) a[i][j] = (s[j] == 'x');
			}
			bool fir = 0;
			try(i,1,hang)
			{
				ak = scanf("%s",s+1);
				try(j,1,lie)
				{
					register int x,y;
					if(s[j] == 'x')
					{
						if(!fir) x = i,y = j,fir = 1;
						d[++zhi] = xin_data(i-x+1,j-y+1);
					}
				}
			}
			if(zhi == 0) {printf("No\n"); continue;}
			bool ok = 1;
			try(i,1,n) if(ok) try(j,1,m)
			{
				if(a[i][j])
				{
					if(!check(i,j)) {puts("No"); ok = 0; break;}
				}
			}
			if(ok) puts("Yes");
		}
		return 0;
	}
}
signed main(){return xin::main();}

客星璀璨之夜

直观的暴力就是先枚举排列,然后再去枚举每一个方向,之后再 \(check\)

我因为自己能够搞掉 \(n=10\) 的点,然而跑了 \(10min\)

正解首先我们推出来概率,之后再乘上每一段的 \(x_i - x_{i-1}\)

但是细节挺多的。。



#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define int long long
namespace xin_io
{
	#define scanf nb = scanf
	#define debug cout<<"debug"<<endl
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<2,stdin),p1 == p2) ? EOF : *p1++
	char buf[1<<20],*p1 = buf,*p2 = buf,output[100]; FILE *xinnb;int nb;typedef long long ll; typedef unsigned long long ull;
	void openfile() {xinnb = freopen("t.txt","r",stdin);} void outfile() {xinnb = freopen("o.txt","w",stdout);}
	inline int get()
	{
		register int s = 0,f = 1; register char ch = gc();	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
		while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc();return s * f;
	}
	template<typename type>inline void write(type x,char out)
	{
		if(!x) return putchar('0'),putchar(out),void(); if(x < 0) putchar('-'),x = -x;
		register int cnt = 0;while(x) output[++cnt] = x % 10,x /= 10; 
		throw(i,cnt,1) putchar(output[i] xor 48);return putchar(out),void();
	}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7,mod = 998244353; static const ll llinf = 1e17+7;
namespace xin
{
	inline int ksm(int x,int y)
	{
		register int ret = 1;
		while(y)
		{
			if(y & 1) ret = ret * x % mod;
			x = x * x % mod; y >>= 1;
		}
		return ret % mod;
	}
	int f[maxn*3/1000][maxn*6/1000],n,x[maxn];
	int jsq1,jsq2;
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get();
		try(i,1,((n<<1)|1)) x[i] = get();
		try(i,1,n)
		{
			register int er = ksm((i<<1),mod-2),yi = ksm(i,mod-2);
			try(j,2,(i<<1|1))
			{
				int temp = (((i << 1) | 1) - j) >> 1;
				if(j & 1) (f[i][j] += f[i-1][j] * temp % mod * yi % mod + f[i-1][j-1] * er % mod + f[i-1][j-2] * er % mod + er + (i - temp - 1) * f[i-1][j-2] % mod * yi % mod) %= mod;
				else (f[i][j] += f[i-1][j] * temp % mod * yi % mod + f[i-1][j] *er % mod + er % mod + f[i-1][j-1]  * er % mod + (i - temp - 1) * f[i-1][j-2] % mod * yi) %= mod;
//				cout<<"i = "<<i<<" j = "<<j<<" f[i][j] = "<<f[i][j]<<endl;
			}
		}
		int ans = 0;
		try(i,2,(n<<1 | 1)) (ans += f[n][i]  * ( x[i] - x[i-1])) %= mod; 
		cout<<ans<<endl;
		return 0;
	}
}
signed main(){return xin::main();}

割海成路之日

这个暂时只有 \(25pts\) 垃圾做法。

还没想出来正解。。。

菜爆了。。。

只有 \(25pts\;code\)



#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
namespace xin_io
{
	#define scanf nb = scanf
	#define debug cout<<"debug"<<endl
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<2,stdin),p1 == p2) ? EOF : *p1++
	char buf[1<<20],*p1 = buf,*p2 = buf,output[100]; FILE *xinnb;int nb;typedef long long ll; typedef unsigned long long ull;
	void openfile() {xinnb = freopen("t.txt","r",stdin);} void outfile() {xinnb = freopen("o.txt","w",stdout);}
	inline int get()
	{
		register int s = 0,f = 1; register char ch = gc();	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
		while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc();return s * f;
	}
	template<typename type>inline void write(type x,char out)
	{
		if(!x) return putchar('0'),putchar(out),void(); if(x < 0) putchar('-'),x = -x;
		register int cnt = 0;while(x) output[++cnt] = x % 10,x /= 10; 
		throw(i,cnt,1) putchar(output[i] xor 48);return putchar(out),void();
	}
}
using namespace xin_io; static const int maxn = 3e6+10,inf = 1e9+7; static const ll llinf = 1e17+7;
namespace xin
{
	class xin_edge{public:int next,ver,w;}edge[maxn];
	int head[maxn],cnt = 1;
	inline void add(int x,int y,int w) {edge[++cnt].ver = y; edge[cnt].w = w ; edge[cnt].next = head[x]; head[x] = cnt;}
	int n,m;
	int dep[maxn],f[maxn][24],fr[maxn],lg[maxn],num = 0;
	std::unordered_map<int,int>e[maxn];
	void pre_dfs(int x,int fa)
	{
		dep[x] = dep[fa] + 1;
		f[++num][0] = x;
		if(!fr[x]) fr[x] = num;
		for(register int i=head[x];i;i=edge[i].next)
		{
			register int y = edge[i].ver;
			if(y == fa) continue;
			pre_dfs(y,x);
			f[++num][0] = x;
			if(!fr[x]) fr[x] = num;
		}
	}
	void RMQ()
	{
		for(register int j=1;(1<<j)<=num;++j)
			for(register int i=1;i+(1<<j)-1<=num;++i)
				if(dep[f[i][j-1]]<dep[f[i+(1<<(j-1))][j-1]])f[i][j]=f[i][j-1];
				else f[i][j]=f[i+(1<<(j-1))][j-1];
		try(i,2,num) lg[i] = lg[i>>1] + 1;
	}
	int lca(int x,int y)
	{
		x = fr[x]; y = fr[y];
		if(x > y) x ^= y ^= x ^= y;
		int len=lg[y-x+1];
		return dep[f[x][len]]<dep[f[y-(1<<len)+1][len]]?f[x][len]:f[y-(1<<len)+1][len];
	}
	bool ok = 0,shit = 1;
	void dfs1(int x,int goal)
	{
		if(x == goal or !shit) return ;
		for(register int i=head[x];i;i=edge[i].next)
		{
			register int y = edge[i].ver,z = edge[i].w;
			if(dep[y] > dep[x]) continue;
			if(ok  and z > 1) {shit = 0; return ;}
			if(z == 3) ok = 1;
			dfs1(y,goal);
		}
	}
	void dfs2(int x,int goal)
	{
		if(x == goal ) return ;
		for(register int i=head[x];i;i=edge[i].next)
		{	
			register int y = edge[i].ver,z = edge[i].w;
			if(dep[y] > dep[x]) continue;
			dfs2(y,goal);
			if(ok and z > 1) {shit = 0; return ;}
			if(z == 3) ok = 1;
		}
	}
	inline bool check(int x,int y)
	{
		register int goal = lca(x,y); ok = 0; shit = 1;
		dfs1(x,goal); dfs2(y,goal);
		return shit;
	}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get(); m = get();
		try(i,1,n-1)
		{
			register int x = get(),y = get(),z = get();
			add(x,y,z); e[x][y] = e[y][x] = cnt;
			add(y,x,z);
		}
		pre_dfs(1,0); RMQ();
		try(i,1,m)
		{
			register int x = get(),y = get(),s = get(),t = get(),ans = 0;
			if(edge[e[x][y]].w xor 1) edge[e[x][y]].w--,edge[e[x][y] xor 1].w--;
			write(check(t,s),' ');
			try(i,1,n) if(check(i,s)) {/*cout<<"i = "<<i<<endl; */ans++;}
			write(ans,'\n');
		}
		return 0;
	}
}
signed main(){return xin::main();}

原文地址:https://www.cnblogs.com/NP2Z/p/15084898.html