【NOIp2019模拟】题解

T1

傻逼题，速度的函数是收敛的，一个等比数列求和就完了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define pb push_back
#define re register
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define cs const
cs double pi=acos(-1);
double thi,v,d,g,vx,vy,t;
int main(){
	int T=read();
	while(T--){
		scanf("%lf%lf%lf%lf",&thi,&v,&d,&g);
		thi=thi/180.0;
		vx=v*cos(pi*thi),vy=v*sin(pi*thi);
		t=vy/(1.0-d*d),t/=g,t=t*2;
		printf("%.5lf
",t*vx);
	}
}

T2：

水题

考虑说可以发现我们要把所有奇数拿出来
考虑说一个数不断乘$2$到$n$有几个数
如果是偶数个就给$2$个集合相同的贡献
奇数个就会有一个集合多$1$

最后要2个集合分别是$m$和$n-m$
选数的方案就是一个组合数

用卢卡斯定理做一下就完了

结果无解的情况少判了一个为负
挂成$20pts$

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
#define int long long
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define pb push_back
#define re register
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define cs const
cs int mod=10000019;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(res=mul(res,a));return res;}
int fac[mod+100],ifac[mod+100];
int n,q,m;
inline void init(){
	fac[0]=ifac[0]=1;
	for(int i=1;i<mod;i++)fac[i]=mul(fac[i-1],i);
	ifac[mod-1]=ksm(fac[mod-1],mod-2);
	for(int i=mod-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline int C(int n,int m){
	return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m]));
}
inline int Lucas(int n,int m){
	if(n<m)return 0;
	if(n<mod&&m<mod)return C(n,m);
	return mul(Lucas(n/mod,m/mod),C(n%mod,m%mod));
}
int sip,plx;
inline int howmany(int l,int r){
	int del=r-l+1;
	if(del&1)return (del/2)+((l&1)==1);
	return del/2;
}
inline void divid(int x){
	int r=x,l,cur=1;
	while(r){
		l=r/2+1;
		if(cur)sip+=howmany(l,r);
		else plx+=howmany(l,r);
		r=l-1,cur^=1;
	}
}
signed main(){
	init();
	n=read(),q=read();
	divid(n);
	int bas=ksm(2,plx);
	while(q--){
		m=read();
		int del=sip+m+m-n;
		if(del&1){cout<<0<<'
';continue;}
		if(del<0){cout<<0<<'
';continue;} 
		del/=2;
		cout<<mul(bas,Lucas(sip,del))<<'
';
	}
}

T3：

傻逼题
和二分图差不多，对于连断每次乘逆元就可以了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define pb push_back
#define re register
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
#define cs const
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(res=mul(res,a));return res;}
cs int N=100005;
int n,m,ans;
int inv[N],anc[N];
inline void init(){
	inv[0]=inv[1]=1;
	for(int i=2;i<=n;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
}
namespace Set{
	int fa[N],siz[N],top;
	pii stk[N];
	inline void init(){
		for(int i=1;i<=n;i++)fa[i]=i,siz[i]=1;
	}
	inline int find(int x){
		return fa[x]==x?x:find(fa[x]);
	}
	inline void merge(int u,int v){
		int f1=find(u),f2=find(v);
		if(f1!=f2){
			if(siz[f1]>siz[f2])swap(f1,f2);
			Mul(ans,inv[siz[f1]]),Mul(ans,inv[siz[f2]]);
			fa[f1]=f2,siz[f2]+=siz[f1];
			Mul(ans,siz[f2]);
			stk[++top]=pii(f1,f2);
		}
	}
	inline void getback(int pre){
		static int u,v;
		while(top>pre){
			u=stk[top].fi,v=stk[top].se;
			Mul(ans,inv[siz[v]]);
			fa[u]=u,siz[v]-=siz[u];
			Mul(ans,siz[u]),Mul(ans,siz[v]);
			top--;
		}
	}
}
map<pii,int> tt;
vector<pii> e[N<<2];
#define lc (u<<1)
#define rc ((u<<1)|1)
#define mid ((l+r)>>1)
void update(int u,int l,int r,int st,int des,pii k){
	if(st<=l&&r<=des){e[u].pb(k);return;}
	if(st<=mid)update(lc,l,mid,st,des,k);
	if(mid<des)update(rc,mid+1,r,st,des,k);
}
void dfs(int u,int l,int r){
	int pre=Set::top;
	for(int i=0;i<e[u].size();i++){
		pii x=e[u][i];
		Set::merge(x.fi,x.se);
	}
	if(l==r){anc[l]=ans;}
	else dfs(lc,l,mid),dfs(rc,mid+1,r);
	Set::getback(pre);
}
vector<pii> E;
int main(){
	n=read(),m=read();
	Set::init();
	init(),ans=1;
	for(int i=1;i<=m;i++){
		int op=read();
		if(op==1){
			int u=read(),v=read();
			if(u>v)swap(u,v);
			tt[pii(u,v)]=i;
			E.pb(pii(u,v));
		}
		else{
			int u=read(),v=read();
			if(u>v)swap(u,v);
			update(1,1,m,tt[pii(u,v)],i-1,pii(u,v));
			tt.erase(pii(u,v));
		}
	}
	for(int i=0;i<E.size();i++){
		pii x=E[i];
		if(tt.count(x))update(1,1,m,tt[x],m,x),tt.erase(x);
	}
	dfs(1,1,m);
	for(int i=1;i<=m;i++)cout<<anc[i]<<"
";
}