【2020省选模拟】题解
T1:
考虑设表示前个走到第二个位置的步数,表示走到第三个位置的步数
然后可以发现
然后优化即可
没有循环展开跑的贼慢
#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs 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++;
}
inline int read(){
char ch=gc();
int res=0;bool 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;
}
inline ll readll(){
char ch=gc();
ll res=0;bool 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;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(int x){return (x<0)?x+mod:x;}
cs int N=1025,C=1023;
struct mat{
int a[3][3];
mat(){memset(a,0,sizeof(a));}
inline void init(){
a[0][0]=0,a[0][1]=1,a[0][2]=0,a[1][0]=2,a[1][1]=2,a[1][2]=0,a[2][0]=1,a[2][1]=2,a[2][2]=1;
}
friend inline mat operator *(cs mat &a,cs mat &b){
mat c;
for(int i=0;i<3;i++)
for(int k=0;k<3;k++)
for(int j=0;j<3;j++)
Add(c.a[i][j],mul(a.a[i][k],b.a[k][j]));
return c;
}
}bas,pw1[N],pw2[N],pw3[N],pw4[N],I;
int main(){
bas.init();I.a[0][0]=1,I.a[1][1]=1,I.a[2][2]=1;
pw1[0]=I;
for(int i=1;i<N;i++)pw1[i]=pw1[i-1]*bas;
bas=pw1[N-1];
pw2[0]=I;
for(int i=1;i<N;i++)pw2[i]=pw2[i-1]*bas;
bas=pw2[N-1];
pw3[0]=I;
for(int i=1;i<N;i++)pw3[i]=pw3[i-1]*bas;
bas=pw3[N-1];
pw4[0]=I;
for(int i=1;i<N;i++)pw4[i]=pw4[i-1]*bas;
int res1=0,res2=0;
int T=read();
while(T--){
ll x=readll();
int c1=(x&C),c2=(x>>10)&C,c3=(x>>20)&C,c4=(x>>30)&C;
mat now=pw1[c1]*pw2[c2]*pw3[c3]*pw4[c4];
res1^=now.a[2][0],res2^=now.a[2][1];
}
cout<<res1<<" "<<res2<<'\n';
return 0;
}
T2:
首先显然就是要给几个组分配数字,让每个组内排成环,相邻乘积之和最大
首先由个环
显然每个组取值相邻的一段最优
每一个组按这样排最优
然后做完了
#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs 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++;
}
inline int read(){
char ch=gc();
int res=0;bool 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;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
int gcd(int a,int b){
return b?gcd(b,a%b):a;
}
cs int N=5005;
int n,m,a[N],b[N];
inline ll calc(int *a,int n){
ll res=0;int cnt=0;
for(int i=1;i<=n;i+=2)b[++cnt]=a[i];
for(int i=n-(n&1);i;i-=2)b[++cnt]=a[i];
for(int i=1;i<n;i++)res+=1ll*b[i]*b[i+1];
res+=1ll*b[n]*b[1];
return res;
}
int main(){
n=read();
for(int i=1;i<=n;i++)a[i]=read();
sort(a+1,a+n+1);
m=read();
while(m--){
int k=read(),t=gcd(k,n),len=n/t;
ll res=0;
for(int i=1;i<=t;i++)res+=calc(a+len*(i-1),len);
cout<<res<<'\n';
}return 0;
}
T3:
艹这是我之前想出来打算做题的一个
显然任意两人之间贡献独立
考虑对于两人
若对于求的概率
那么显然要分成两部分来算
第一部分贡献相当于是一个等差数列,第二部分是直接乘一个
第二部分直接线段树维护
第一部分可以用线段树维护每个位置的系数与贡献
比如记录的,系数就是
然后大概推一下怎么算即可
#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs 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++;
}
inline int read(){
char ch=gc();
int res=0;bool 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;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005,lim=100000;
typedef pair<double,double> pp;
namespace t1{
cs int N=::N<<2;
double s1[N],s2[N],coef[N],tag[N];
#define lc (u<<1)
#define rc ((u<<1)|1)
#define mid ((l+r)>>1)
inline void pushup(int u){
s1[u]=s1[lc]+s1[rc],s2[u]=s2[lc]+s2[rc];
}
void build(int u,int l,int r){
if(l==r){coef[u]=0.5+l;return;}
build(lc,l,mid),build(rc,mid+1,r);
coef[u]=coef[lc]+coef[rc];
}
inline void pushnow(int u,int l,double k){
s1[u]+=k*l,s2[u]+=coef[u]*k,tag[u]+=k;
}
inline void pushdown(int u,int l,int r){
if(tag[u]==0)return;
pushnow(lc,mid-l+1,tag[u]),pushnow(rc,r-mid,tag[u]);
tag[u]=0;
}
void update(int u,int l,int r,int st,int des,double k){
if(st<=l&&r<=des)return pushnow(u,r-l+1,k);
pushdown(u,l,r);
if(st<=mid)update(lc,l,mid,st,des,k);
if(mid<des)update(rc,mid+1,r,st,des,k);
pushup(u);//cout<<u<<" "<<l<<" "<<r<<" "<<st<<" "<<des<<" "<<k<<" "<<s1[u]<<" "<<s2[u]<<'\n';
}
inline pp operator +(cs pp &a,cs pp &b){return pp(a.fi+b.fi,a.se+b.se);}
pair<double,double> query(int u,int l,int r,int st,int des){
if(st>des)return pp(0,0);
if(st<=l&&r<=des)return pp(s1[u],s2[u]);
pushdown(u,l,r);
if(des<=mid)return query(lc,l,mid,st,des);
if(mid<st)return query(rc,mid+1,r,st,des);
return query(lc,l,mid,st,des)+query(rc,mid+1,r,st,des);
}
}
namespace t2{
cs int N=::N<<2;
double s[N],tag[N];
inline void pushup(int u){
s[u]=s[lc]+s[rc];
}
inline void pushnow(int u,int l,double k){
s[u]+=k*l,tag[u]+=k;
}
inline void pushdown(int u,int l,int r){
if(tag[u]==0)return;
pushnow(lc,mid-l+1,tag[u]),pushnow(rc,r-mid,tag[u]);
tag[u]=0;
}
void update(int u,int l,int r,int st,int des,double k){
if(st<=l&&r<=des)return pushnow(u,r-l+1,k);
pushdown(u,l,r);
if(st<=mid)update(lc,l,mid,st,des,k);
if(mid<des)update(rc,mid+1,r,st,des,k);
pushup(u);
}
double query(int u,int l,int r,int st,int des){
if(st<=l&&r<=des)return s[u];
double res=0;pushdown(u,l,r);
if(st<=mid)res+=query(lc,l,mid,st,des);
if(mid<des)res+=query(rc,mid+1,r,st,des);
return res;
}
#undef lc
#undef rc
#undef mid
}
int n,l[N],r[N];
int main(){
n=read();
for(int i=1;i<=n;i++)l[i]=read(),r[i]=read();
t1::build(1,1,lim);
for(int i=1;i<=n;i++)t1::update(1,1,lim,l[i],r[i]-1,1.0/(r[i]-l[i])),t2::update(1,1,lim,l[i],r[i]-1,1.0/(r[i]-l[i]));//,cout<<l[i]<<" "<<r[i]-1<<" "<<(1.0/(r[i]-l[i]))<<'\n';
for(int i=1;i<=n;i++){
pp x=t1::query(1,1,lim,l[i],r[i]-1);
double c=t2::query(1,1,lim,r[i],lim);
double res=x.se-x.fi*l[i]+c*(r[i]-l[i]);res/=(r[i]-l[i]),res+=0.5;
printf("%.8lf\n",res);
}return 0;
}