2024初三集训模拟测试3

2024初三集训模拟测试3

  1. T1 排序：

    显然贪心。

    将 1ll*a[i]*a[i-1] \(\to\) 1ll*(a[i]*a[i-1]) 囍爆零

    CODE 
    #include<bits/stdc++.h>
using namespace std;
using llt=long long;
using ull=unsigned long long;
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
const int N=1e5+3;
int n,a[N<<2];

namespace IO{
    template<class T> inline void write(T x){
        static T st[45];T top=0;if(x<0)x=-x,putchar('-');
        do{st[top++]=x%10;}while(x/=10);while(top)putchar(st[--top]^48);
    }
    template<class T> T READ_NONPARAMETRIC_INIT;
    template<class T = int> inline T read(T &x=READ_NONPARAMETRIC_INIT<T>){
        char s=getchar();x=0;bool pd=false;while(s<'0'||'9'<s){if(s=='-') pd=true;s=getchar();}
        while('0'<=s&&s<='9'){x=x*10+(s^48),s=getchar();} return (pd?(x=-x):x);
    }
}
namespace IO{
    char NL_C; double NL_F; long double NL_LF;
    inline char read(char &c){c=getchar();while(c<33||c>126) c=getchar();return c;}
    template<int MAXSIZE=INT_MAX> inline int read(char* c){
        char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
        while(s>=33&&s<=126&&pos<MAXSIZE) c[pos++]=s,s=getchar();c[pos]='\0';return pos;
    }
    template<int MAXSIZE=INT_MAX> inline int read(string &c){
        c.clear();char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
        while(s>=33&&s<=126&&pos<MAXSIZE) c.push_back(s),s=getchar(),pos++;return pos;
    }
    inline double read(double &x){scanf("%lf",&x);return x;}
    inline long double read(long double &x){scanf("%Lf",&x);return x;}
    template<class T,class... Args> inline void read(T& x,Args&... args){read(x);read(args...);}
    inline void write(char c){putchar(c);}
    inline void write(char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
    inline void write(string &c){int len=c.size();For(i,0,len-1,1) putchar(c[i]);}
    inline void write(const char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
    template<int PRECISION=6> inline void write(double x){printf("%.*lf",PRECISION,x);}
    template<int PRECISION=6> inline void write(long double x){printf("%.*Lf",PRECISION,x);}
    template<class T> inline void Write(T x){write(x),putchar(' ');}
    inline void Write(char c){write(c);if(c!='\n') putchar(' ');}
    inline void Write(char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
    inline void Write(string &c){write(c);if(c[c.size()-1]!='\n') putchar(' ');}
    inline void Write(const char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
    template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
    template<class T,class... Args> inline void Write(T x,Args... args){Write(x);Write(args...);}
}
using namespace IO;

int main(){
#ifndef ONLINE_JUDGE
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
#endif
    read(n);
    For(i,1,n<<2,1) read(a[i]);
    sort(a+1,a+(n<<2)+1);
    llt ans=0;
    For(i,1,n,1) ans-=1ll*a[i]*a[(n<<1)-i+1];
    For_(i,n<<2,(n<<1)+1,2) ans+=1ll*a[i]*a[i-1];
    write(ans);
}

  2. T2 牛吃草：

    显然二分答案。

    考虑 \(dp\) 验证。

    设 \(dp[i][1/0]\) 表示在前 \(i\) 个选、不选的方案数。

    显然转移。

    但是要单调队列优化，

    因为有个常数不好转移，可以同一加上 \(n-i\)，再在查询时减掉。

    CODE 
    #include<bits/stdc++.h>
using namespace std;
using llt=long long;
using ull=unsigned long long;
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
const int N=5e5+3;
int n,w[N],s,dp[N][2],que[N],hd,tl=1;

namespace IO{
    template<class T> inline void write(T x){
        static T st[45];T top=0;if(x<0)x=-x,putchar('-');
        do{st[top++]=x%10;}while(x/=10);while(top)putchar(st[--top]^48);
    }
    template<class T> T READ_NONPARAMETRIC_INIT;
    template<class T = int> inline T read(T &x=READ_NONPARAMETRIC_INIT<T>){
        char s=getchar();x=0;bool pd=false;while(s<'0'||'9'<s){if(s=='-') pd=true;s=getchar();}
        while('0'<=s&&s<='9'){x=x*10+(s^48),s=getchar();} return (pd?(x=-x):x);
    }
}
namespace IO{
    char NL_C; double NL_F; long double NL_LF;
    inline char read(char &c){c=getchar();while(c<33||c>126) c=getchar();return c;}
    template<int MAXSIZE=INT_MAX> inline int read(char* c){
        char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
        while(s>=33&&s<=126&&pos<MAXSIZE) c[pos++]=s,s=getchar();c[pos]='\0';return pos;
    }
    template<int MAXSIZE=INT_MAX> inline int read(string &c){
        c.clear();char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
        while(s>=33&&s<=126&&pos<MAXSIZE) c.push_back(s),s=getchar(),pos++;return pos;
    }
    inline double read(double &x){scanf("%lf",&x);return x;}
    inline long double read(long double &x){scanf("%Lf",&x);return x;}
    template<class T,class... Args> inline void read(T& x,Args&... args){read(x);read(args...);}
    inline void write(char c){putchar(c);}
    inline void write(char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
    inline void write(string &c){int len=c.size();For(i,0,len-1,1) putchar(c[i]);}
    inline void write(const char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
    template<int PRECISION=6> inline void write(double x){printf("%.*lf",PRECISION,x);}
    template<int PRECISION=6> inline void write(long double x){printf("%.*Lf",PRECISION,x);}
    template<class T> inline void Write(T x){write(x),putchar(' ');}
    inline void Write(char c){write(c);if(c!='\n') putchar(' ');}
    inline void Write(char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
    inline void Write(string &c){write(c);if(c[c.size()-1]!='\n') putchar(' ');}
    inline void Write(const char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
    template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
    template<class T,class... Args> inline void Write(T x,Args... args){Write(x);Write(args...);}
}
using namespace IO;
namespace DQ{
    struct Q{int t,id;}que[N];
    int hd=1,tl=0;
    inline void Clr(){hd=1,tl=0;}
    inline void Add(int x,int id){
        while(hd<=tl&&que[tl].t<x) tl--;
        que[++tl]={x,id};
    }
    inline int Get(int l){
        while(hd<=tl&&que[hd].id<l) hd++;
        if(hd>tl) return 0;
        return que[hd].t;
    }
}

inline bool check(int sz){
    memset(dp,0,sizeof dp); DQ::Clr();
    For(i,1,n,1){
        if(i>=sz) DQ::Add(max(dp[i-sz][1],dp[i-sz][0])+n-(i-sz),i-sz);
        dp[i][0]=max(dp[i-1][1],dp[i-1][0]);
        dp[i][1]=max(dp[i][1],DQ::Get(i-w[i])+i-n);
    }
    if(max(dp[n][0],dp[n][1])<s) return 0;
    else return 1;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
#endif
    read(n);
    For(i,1,n,1) read(w[i]);
    read(s);s=ceil(n*1.0/100*s);
    int l=1,r=n;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid)) l=mid+1;
        else r=mid-1;
    }
    write(r);
}

  3. T3 树上的宝藏：

    考虑先不更换，显然树形 DP。

    因为只能换一条，可以考虑换根，枚举到 \(u\) 子节点 \(v\) 时转移。

    这里用了 \(dp_{0/1}\) 表示不更换以 \(1\) 为根选、不选的方案数，\(s_{u,0/1}\) 表示不更换以 \(u\) 为根选、不选的方案，\(ans_{u,0/1}\) 表示将 \(fa_u \to u\) 替换后选、不选的方案。

    可以少用一个，但无所谓。

    CODE 
    #include<bits/stdc++.h>
using namespace std;
using llt=long long;
using ull=unsigned long long;
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
const int N=3e5+4,MOD=998244353;
int n,dp[N][2],ans[N][2],fa[N],s[N][2];
struct DG{int x,y;}nw;
queue<DG> que;

namespace IO{
	template<class T> inline void write(T x){
		static T st[45];T top=0;if(x<0)x=-x,putchar('-');
		do{st[top++]=x%10;}while(x/=10);while(top)putchar(st[--top]^48);
	}
	template<class T> T READ_NONPARAMETRIC_INIT;
	template<class T = int> inline T read(T &x=READ_NONPARAMETRIC_INIT<T>){
		char s=getchar();x=0;bool pd=false;while(s<'0'||'9'<s){if(s=='-') pd=true;s=getchar();}
		while('0'<=s&&s<='9'){x=x*10+(s^48),s=getchar();} return (pd?(x=-x):x);
	}
}
namespace IO{
	char NL_C; double NL_F; long double NL_LF;
	inline char read(char &c){c=getchar();while(c<33||c>126) c=getchar();return c;}
	template<int MAXSIZE=INT_MAX> inline int read(char* c){
		char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
		while(s>=33&&s<=126&&pos<MAXSIZE) c[pos++]=s,s=getchar();c[pos]='\0';return pos;
	}
	template<int MAXSIZE=INT_MAX> inline int read(string &c){
		c.clear();char s=getchar();int pos=0;while(s<33||s>126) s=getchar();
		while(s>=33&&s<=126&&pos<MAXSIZE) c.push_back(s),s=getchar(),pos++;return pos;
	}
	inline double read(double &x){scanf("%lf",&x);return x;}
	inline long double read(long double &x){scanf("%Lf",&x);return x;}
	template<class T,class... Args> inline void read(T& x,Args&... args){read(x);read(args...);}
	inline void write(char c){putchar(c);}
	inline void write(char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
	inline void write(string &c){int len=c.size();For(i,0,len-1,1) putchar(c[i]);}
	inline void write(const char *c){int len=strlen(c);For(i,0,len-1,1) putchar(c[i]);}
	template<int PRECISION=6> inline void write(double x){printf("%.*lf",PRECISION,x);}
	template<int PRECISION=6> inline void write(long double x){printf("%.*Lf",PRECISION,x);}
	template<class T> inline void Write(T x){write(x),putchar(' ');}
	inline void Write(char c){write(c);if(c!='\n') putchar(' ');}
	inline void Write(char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
	inline void Write(string &c){write(c);if(c[c.size()-1]!='\n') putchar(' ');}
	inline void Write(const char *c){write(c);if(c[strlen(c)-1]!='\n') putchar(' ');}
	template<class T,class... Args> inline void write(T x,Args... args){write(x);write(args...);}
	template<class T,class... Args> inline void Write(T x,Args... args){Write(x);Write(args...);}
}
using namespace IO;
namespace MT{
	inline int Fpw(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;
	}
	inline int Inv(int a){return Fpw(a%MOD,MOD-2);}
}using MT::Inv;
namespace TO{
	int hd[N],nt[N<<1],to[N<<1],tot;
	inline void Add(int u,int v){to[++tot]=v,nt[tot]=hd[u],hd[u]=tot;}
	#define For_to(i,u,v) for(int i=TO::hd[u],v=TO::to[i];i;i=TO::nt[i],v=TO::to[i])
}
inline void Dfs1(int u,int f){
	fa[u]=f,dp[u][0]=dp[u][1]=1;
	For_to(i,u,v) if(v!=f){
		Dfs1(v,u);
		dp[u][1]=1ll*dp[v][0]*dp[u][1]%MOD;
		dp[u][0]=1ll*(dp[v][1]+dp[v][0])*dp[u][0]%MOD;
	}
}
inline int Get0(int u){return 1ll*s[fa[u]][0]*Inv(dp[u][1]+dp[u][0])%MOD;}
inline int Get1(int u){return 1ll*s[fa[u]][1]*Inv(dp[u][0])%MOD;}
inline void Dfs2(int u,int f){
	if(f){
		s[u][0]=(Get0(u)+Get1(u))%MOD;
		s[u][1]=(Get0(u))%MOD;
	}else s[u][0]=s[u][1]=1;
	For_to(i,u,v) if(v!=f){
		s[u][0]=1ll*s[u][0]*(dp[v][0]+dp[v][1])%MOD;
		s[u][1]=1ll*s[u][1]*dp[v][0]%MOD;
	}
	For_to(i,u,v) if(v!=f){
		ans[v][0]=1ll*Get1(v)*dp[v][0]%MOD;
		ans[v][1]=1ll*(Get0(v)+Get1(v))*dp[v][1]%MOD;
		Dfs2(v,u);
	}
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in_out/in.in","r",stdin);
	freopen("in_out/out.out","w",stdout);
#endif
	read(n);
	For(i,1,n-1,1){
		int u,v;read(u,v);que.push(DG{u,v});
		TO::Add(u,v),TO::Add(v,u);
	}
	Dfs1(1,0);
	Dfs2(1,0);
	while(!que.empty()){
		int u=que.front().x,v=que.front().y;que.pop();
		if(fa[u]==v) swap(u,v);
		write((ans[v][0]+ans[v][1])%MOD,'\n');
	}
}

  4. T4 MEX：

    不会，先咕了。

posted @ 2024-02-22 21:06  xrlong  阅读(30)  评论(0)  编辑  收藏  举报