2019 牛客多校第九场

A The Power Of Fobonacci

B Quadratic equation

参考代码：

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
typedef long long LL;
#define int long long
int quick_pow(int a,int b,int p=mod)
{
    int ans=1;
    while(b)
    {
        if(b&1) ans=1LL*ans*a%mod;
        a=1LL*a*a%mod;
        b>>=1;
    }
    return ans;
}
namespace Cipolla{
    LL w,P;
    struct E{
        LL x,y;
        E(LL _x,LL _y):x(_x),y(_y){}
        friend E operator *(E a,E b){
            return E((a.x*b.x%P+a.y*b.y%P*w)%P,(a.x*b.y%P+a.y*b.x%P)%P);
        }
    };
    inline E Pow(E x,int y){
        E ret=E(1,0);
        for(;y;y>>=1,x=x*x) if(y&1) ret=ret*x;
        return ret;
    }
    LL calc(LL x,LL p){
        P=p; x%=P;
        if(x==0) return 0;
        LL tmp=quick_pow(x,(p-1)/2,p);
        if(tmp!=1) return -1;
        while(1){
            tmp=rand()%p;
            LL cur=(tmp*tmp%P+P-x)%P;
            if(quick_pow(cur,(p-1)/2,p)==p-1) break;
        }
        w=(tmp*tmp%P+P-x)%P;
        E A=E(tmp,1);
        return Pow(A,(p+1)/2).x;
    }
}
 
int32_t main()
{
    int t;
    scanf("%lld",&t);
    int b,c;
    int inv2=(mod+1)/2;
    while(t--)
    {
        scanf("%lld%lld",&b,&c);
        int delta=Cipolla::calc(b*b-4LL*c+mod,mod)%mod;
        if(delta==-1)
        puts("-1 -1");
        else
        {
            int y=(1LL*(b+delta)*inv2%mod+mod)%mod;
            int x=(1LL*(b-delta)*inv2%mod+mod)%mod;
            if(x>y) swap(x,y);
            printf("%lld %lld
",x,y);
        }
    }
}      

C  Inversions of all permutations

D Knapsack Cryptosystem

题解：https://blog.csdn.net/liufengwei1/article/details/99650612

#include<bits/stdc++.h>
#define maxl 20
using namespace std;
 
int n,m,tot,cnt;
int dy[(1<<maxl)];
long long ss,ans;
long long a[maxl],b[maxl];
long long suma[1<<maxl],sumb[1<<maxl];
struct node
{
    long long num;
    long long val;
    bool operator < (const node &b)const
    {
        return val<b.val;
    }
}c[1<<maxl];
 
inline void prework()
{
    n=tot/2;m=tot-n;
    for(int i=n;i>=1;i--)
        scanf("%lld",&a[i]);
    for(int i=m;i>=1;i--)
        scanf("%lld",&b[i]);
}
 
inline bool cmp(const node &x,const node &y)
{
    return x.val<y.val;
}
 
inline void mainwork()
{
    if(ss==0)
    {
        ans=0;
        return;
    }
    suma[0]=0;sumb[0]=0;
    cnt=0;
    c[++cnt]=node{0,0};
    int up=1<<n,t;
    for(int s=1;s<up;s++)
    {
        t=s-(s&-s);
        suma[s]=suma[t]+a[dy[s&-s]+1];
        c[++cnt]=node{s,suma[s]};
    }
    sort(c+1,c+1+cnt,cmp);
    up=1<<m;int id;
    for(int s=1;s<up;s++)
    {
        t=s-(s&-s);
        sumb[s]=sumb[t]+b[dy[s&-s]+1];
        id=lower_bound(c+1,c+1+cnt,node{s,ss-sumb[s]})-c;
        if(id<=cnt && c[id].val==ss-sumb[s])
        {
            ans=c[id].num*(1<<m)+s;
            break;
        }
    }
}
 
inline void print()
{
    for(int i=tot-1;i>=0;i--)
        printf("%lld",((ans>>i)&1));
    puts("");
}
 
int main()
{
    for(int i=0;i<=19;i++)
        dy[1<<i]=i;
    while(~scanf("%d%lld",&tot,&ss))
    {
        prework();
        mainwork();
        print();
    }
    return 0;
}

E All men are brothers

题解：https://blog.csdn.net/liufengwei1/article/details/99655850

#include<bits/stdc++.h>
#define maxl 100010
using namespace std;
 
int n,m,top;
int f[maxl],tot,s[maxl];
__int128 t[maxl],one=1;
__int128 sum[5];
 
inline void prework()
{
    tot=n;
    for(int i=1;i<=n;i++)
        f[i]=i,t[i]=one;
    sum[1]=one*n;
    sum[2]=one*n*(n-1)/2;
    sum[3]=one*n*(n-1)*(n-2)/6;
    sum[4]=one*n*(n-1)*(n-2)*(n-3)/24;
}
 
inline void print(__int128 x)
{
    top=0;
    if(x==0)
        s[++top]=0;
    while(x>0)
    {
        s[++top]=x%10;
        x/=10;
    }
    for(int i=top;i>=1;i--)
        printf("%d",s[i]);
    puts("");
}
 
inline int find(int x)
{
    if(f[x]!=x)
        f[x]=find(f[x]);
    return f[x];
}
 
inline void mainwork()
{
    print(sum[4]);
    int u,v,x,y;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&u,&v);
        x=find(u);y=find(v);
        if(x!=y)
        {
            tot--;
            if(tot>=4)
            {
                sum[1]-=t[x]+t[y];
                sum[2]-=t[x]*sum[1]+t[y]*sum[1]+t[x]*t[y];
                sum[3]-=t[x]*sum[2]+t[y]*sum[2]+t[x]*t[y]*sum[1];
                sum[4]-=t[x]*sum[3]+t[y]*sum[3]+t[x]*t[y]*sum[2];
                t[x]+=t[y];f[y]=x;
                sum[4]+=t[x]*sum[3];
                sum[3]+=t[x]*sum[2];
                sum[2]+=t[x]*sum[1];
                sum[1]+=t[x];
            }
            else
                sum[4]=0;
        }
        print(sum[4]);
    }
}
 
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        prework();
        mainwork();
    }
    return 0;
}

F Birthday Reminders

G Checkers

H Cutting Bamboos

题解：https://blog.csdn.net/liufengwei1/article/details/99655160

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define eps 1e-8
const int maxn=1e5+10;
const int maxm=2e5+10;
int n,q,rt[maxn*40],tot;
ll a[maxm],pre[maxm];
struct Node{
    int ls,rs;
    ll sum,cnt;
} tr[maxn*40];

inline void Insert(int y,int &x,int l,int r,int val)
{
    tr[++tot]=tr[y];tr[tot].sum+=val;++tr[tot].cnt; x=tot;
    if(l==r) return ;
    int mid=l+r>>1;
    if(val<=mid) Insert(tr[y].ls,tr[x].ls,l,mid,val);
    else Insert(tr[y].rs,tr[x].rs,mid+1,r,val);
}

inline double Query(int x,int y,int l,int r,int L,int R,double val)
{
    //cout<<"Q"<<endl;
    double res=0;
    if(l==L&&r==R)
    {
        res=(tr[y].sum-tr[x].sum)-(tr[y].cnt-tr[x].cnt)*val;
        return res;
    }
    int mid=l+r>>1;
    if(R<=mid) res=Query(tr[x].ls,tr[y].ls,l,mid,L,R,val);
    else if(L>mid) res=Query(tr[x].rs,tr[y].rs,mid+1,r,L,R,val);
    else
    {
        res+=Query(tr[x].ls,tr[y].ls,l,mid,L,mid,val);
        res+=Query(tr[x].rs,tr[y].rs,mid+1,r,mid+1,R,val);
    }
    return res;
}

int main()
{
    scanf("%d%d",&n,&q); tot=0;
    for(int i=1;i<=n;++i) scanf("%lld",a+i),pre[i]=pre[i-1]+a[i];
    tr[rt[0]].sum=tr[rt[0]].cnt=0;tr[rt[0]].ls=0;tr[rt[0]].ls=0;
    for(int i=1;i<=n;++i) Insert(rt[i-1],rt[i],1,maxn,a[i]);
    while(q--)
    {
        int l,r,x,y;
        scanf("%d%d%d%d",&l,&r,&x,&y);
        double L=0,R=1.0*maxn,mid,res;
        double ans=1.0*(pre[r]-pre[l-1])*x/y;
        while(R-L>eps)
        {
            //cout<<"erfen"<<endl;
            mid=(R+L)/2;
            int num=(int)mid;
            if(mid>1.0*num) ++num;
            res=Query(rt[l-1],rt[r],1,maxn,num,maxn,mid);
            if(res<ans) R=mid;
            else L=mid;
        }
        printf("%.8lf
",L);
    }
    return 0;
}

I KM and M

J Symmetrical Painting

题解：https://blog.csdn.net/liufengwei1/article/details/99657287

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int size=3e5+5;
const int lens=1e6+5;
int sum,tot;
int L[size],R[size];
int idl[size],idr[size],idmid[size];
int cnt[lens][3];
int v[lens];
int main()
{
    int n;
    scanf("%d",&n);
    int l,r;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&L[i],&R[i]);
        L[i]=L[i]*2;
        R[i]=R[i]*2;
    }
    sum=0;
    for(int i=1;i<=n;i++)
    {
        v[++sum]=L[i];
        v[++sum]=R[i];
        v[++sum]=(int)((1ll*L[i]+1ll*R[i])/2);
    }
    sort(v+1,v+1+sum);
    tot=unique(v+1,v+1+sum)-v-1;
    int sz=tot,d;
    for(int i=1;i<=n;i++)
    {
        idl[i]=lower_bound(v+1,v+1+tot,L[i])-v;
        idr[i]=lower_bound(v+1,v+1+tot,R[i])-v;
        d=(int)((1ll*L[i]+1ll*R[i])/2);
        idmid[i]=lower_bound(v+1,v+1+tot,d)-v;
    }
  
    for(int i=1;i<=n;i++)
    {
        cnt[idl[i]][0]++;cnt[idr[i]][1]++;cnt[idmid[i]][2]++;
    }
    LL ty0=cnt[1][0],ty1=cnt[1][1],ty2=cnt[1][2];
    LL ans=0;
    LL tmp=ans;
    for(int i=2;i<=sz;i++)
    {
        LL r=v[i],l=v[i-1];
        tmp=tmp+1LL*(r-l)*(ty0-2*ty2+ty1);
        ty0+=cnt[i][0],ty1+=cnt[i][1],ty2+=cnt[i][2];
//      cout<<ty0<<' '<<ty1<<' '<<ty2<<endl;
//      cout<<tmp<<endl;
        ans=max(ans,tmp);
    }
    printf("%lld
",ans);
    return 0;
}