nowcoder练习赛28

　　https://www.nowcoder.com/acm/contest/200#question

　　最近突然找到了打比赛的乐趣,于是参加了这场比赛.

　　

　　生日宴会:https://www.nowcoder.com/acm/contest/200/A

　　按照题意模拟,没什么好说的.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <cmath>
 # include <algorithm>
 # include <string>
 # define R register int

 using namespace std;

 const int maxn=;
 char name[maxn][];
 int n,m,dat,x,y;
 struct nod
 {
     int val,k;
 }id[][maxn];
 int h[];

 bool cmp (nod a,nod b)
 {
     return a.val<b.val;
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for (R i=;i<=n;++i)
     {
         scanf("%s",name[i]);
         scanf("%d",&dat);
         id[dat%][ ++h[dat%] ].k=i;
         id[dat%][ h[dat%] ].val=dat/;
     }
     for (R i=;i<=;++i)
         sort(id[i]+,id[i]++h[i],cmp);
     for (R i=;i<=m;++i)
     {
         scanf("%d%d",&x,&y);
         printf("%s\n",name[ id[y][x].k ]);
     }
     return ;
 }

A

　

　　数据结构:https://www.nowcoder.com/acm/contest/200/B

　　题意概述:维护一个区间,支持区间加,区间乘,查询区间和,区间平方和.

　　好麻烦的题目啊,本来要写线段树的,后来觉得分块可能简单点就写了分块.事实上...并没有简单,还是要打很多标记,维护起来差不多麻烦.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <cmath>
 # include <algorithm>
 # include <string>
 # define R register int
 # define ll long long

 using namespace std;

 const int maxn=;
 long long x,a[maxn],s[maxn],ss[maxn],delta_a[maxn],delta_b[maxn];
 int b[maxn],n,m,siz,l,r,opt,si[maxn];

 ll read()
 {
     long long x=,f=;
     char c=getchar();
     while (!isdigit(c)) { if(c=='-') f=-f; c=getchar(); }
     while (isdigit(c)) { x=(x<<)+(x<<)+(c^); c=getchar(); }
     return x*f;
 }

 void down (int x)
 {
     s[x]=;
     ss[x]=;
     for (R i=+siz*(x-);i<=min(x*siz,n);++i)
     {
         a[i]=a[i]*delta_a[x]+delta_b[x];
         s[x]+=a[i];
         ss[x]+=a[i]*a[i];
     }
     delta_a[x]=;
     delta_b[x]=;
 }

 long long ask1 (int l,int r)
 {
     long long ans=;
     int id;
     if(b[l]==b[r])
     {
         id=b[l];
         down(id);
         for (R i=l;i<=r;++i) ans+=a[i];
     }
     else
     {
         id=b[l];
         down(id);
         for (R i=l;i<=siz*b[l];++i) ans+=a[i];
         for (R i=b[l]+;i<=b[r]-;++i) ans+=s[i];
         id=b[l];
         down(id);
         for (R i=+siz*(b[r]-);i<=r;++i) ans+=a[i];
     }
     return ans;
 }

 long long ask2 (int l,int r)
 {
     int id;
     long long ans=;
     if(b[l]==b[r])
     {
         id=b[l];
         down(id);
         for (R i=l;i<=r;++i) ans+=a[i]*a[i];
     }
     else
     {
         id=b[l];
         down(id);
         for (R i=l;i<=siz*b[l];++i) ans+=a[i]*a[i];
         for (R i=b[l]+;i<=b[r]-;++i) ans+=ss[i];
         id=b[l];
         down(id);
         for (R i=+siz*(b[r]-);i<=r;++i) ans+=a[i]*a[i];
     }
     return ans;
 }

 void add (int l,int r,ll c)
 {
     int id;
     if(b[l]==b[r])
     {
         id=b[l];
         down(id);
         for (R i=l;i<=r;++i)
         {
             s[id]+=c;
             ss[id]+=c*c+*c*a[i];
             a[i]=a[i]+c;
         }
     }
     else
     {
         id=b[l];
         down(id);
         for (R i=l;i<=siz*b[l];++i)
         {
             s[id]+=c;
             ss[id]+=c*c+*c+a[i];
             a[i]=a[i]+c;
         }
         for (R i=b[l]+;i<=b[r]-;++i)
         {
             delta_a[i]+=c;
             ss[i]+=c*c*si[i]+*c*s[i];
             s[i]+=si[i]*c;
         }
         id=b[r];
         down(id);
         for (R i=+siz*(b[r]-);i<=r;++i)
         {
             s[id]+=c;
             ss[id]+=c*c+*c+a[i];
             a[i]=a[i]+c;
         }
     }
 }

 void mul (int l,int r,ll c)
 {
     int id;
     if(b[l]==b[r])
     {
         id=b[l];
         down(id);
         for (R i=l;i<=r;++i)
         {
             s[id]+=a[i]*(c-);
             ss[id]+=(c*c-)*(a[i]*a[i]);
             a[i]*=c;
         }
     }
     else
     {
         id=b[l];
         down(id);
         for (R i=l;i<=siz*b[l];++i)
         {
             s[id]+=a[i]*(c-);
             ss[id]+=(c*c-)*(a[i]*a[i]);
             a[i]*=c;
         }
         for (R i=b[l]+;i<=b[r]-;++i)
         {
             delta_b[i]*=c;
             delta_a[i]*=c;
             ss[i]*=c;
             s[i]*=c;
         }
         id=b[r];
         down(id);
         for (R i=+siz*(b[r]-);i<=r;++i)
         {
             s[id]+=a[i]*(c-);
             ss[id]+=(c*c-)*(a[i]*a[i]);
             a[i]*=c;
         }
     }
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for (R i=;i<=n;++i) a[i]=read();
     siz=sqrt(n);
     for (R i=;i<=n;++i) b[i]=(i-)/siz+;
     for (R i=;i<=n;++i) s[ b[i] ]+=a[i],ss[ b[i] ]+=a[i]*a[i];
     for (R i=;i<=n;++i) si[ b[i] ]=siz,delta_a[i]=;
     si[ b[n] ]=n-siz*(b[n]-);
     for (R i=;i<=m;++i)
     {
         scanf("%d",&opt);
         if(opt==||opt==)
         {
             scanf("%d%d",&l,&r);
             if(opt==) printf("%lld\n",ask1(l,r));
             else printf("%lld\n",ask2(l,r));
         }
         else
         {
             scanf("%d%d%lld",&l,&r,&x);
             if(opt==) mul(l,r,x);
             else add(l,r,x);
         }
     }
     return ;
 }

B

　　迎风舞:https://www.nowcoder.com/acm/contest/200/E

　　题意概述:物理题.

　　三分抛出方向与水平方向的夹角,解方程算出时间和落地时间即可.关于这个函数为什么是单峰的,这里提供一个不严谨的证明:首先直觉上平抛肯定是比不上上抛的,但是角度为$90$度的上抛运动等于没有抛出去,所以函数应该是满足三分性的.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <cmath>
 # include <algorithm>
 # include <string>
 # define R register int

 using namespace std;

 const double g=9.80665;
 const double eps=0.000001;
 double h,v,l,r,lmid,rmid,ans,lans;
 int T;

 double f (double si)
 {
     double vx,vy,a,b,c,delta,x1,x2,t;
     vy=v*si;
     vx=sqrt(v*v-vy*vy);
     a=-g/2.0;
     b=vy;
     c=h;
     delta=sqrt(b*b-4.0*a*c);

     x1=(-b+delta)/(*a);
     x2=(-b-delta)/(*a);

     t=max(x1,x2);
     return t*vx;
 }

 int main()
 {
     scanf("%d",&T);
     while (T--)
     {
         scanf("%lf%lf",&h,&v);
         l=,r=;
         while (r-l>eps)
         {
             lmid=l+(r-l)/3.0;
             rmid=r-(r-l)/3.0;
             if(f(lmid)<f(rmid))
                 l=lmid;
             else r=rmid;
         }
         printf("%.5lf\n",f(l));
     }
     return ;
 }

E

------------------比赛时就做到这里了,以下是赛后补题------------------

　　随风飘:https://www.nowcoder.com/acm/contest/200/D

　　题意概述:给定$n$个字符串,求它们两两之间的$LCP$的长度和,此外给出$T$组询问$k$,表示有$k$个字符串消失了,求所有的消失情况下$LCP$的长度和的和.

　　其实考试时本来是有机会$A$的,但是都以为开不下存字符串的数组导致纷纷弃疗.开数组靠的是信仰...其实当时为什么不开一个$vector$呢,果然考试时智商会降低.

　　首先把字符串都串到$Trie$树上去,统计$k=0$时的答案.对于其他的询问就不能再上树了!要考虑运用组合数学.每个$LCP$有$C_{n-2}^k$种方法留下来,直接算就好了.

　　

 # include <cstdio>
 # include <iostream>
 # include <cstring>
 # include <cmath>
 # include <algorithm>
 # include <string>
 # define R register int
 # define ll long long
 # define mod 

 using namespace std;

 const int maxs=;
 int k,n,q,len,cnt=,c[][];
 char s[maxs];

 struct Trie
 {
     int ch[maxs][];
     int vis[maxs],s[maxs];
     long long ans;
     void clear()
     {
         ans=;
         memset(ch,,sizeof(ch));
         memset(vis,,sizeof(vis));
         memset(s,,sizeof(s));
     }
     void ins (char *s)
     {
         int len=strlen(s);
         int x=;
         for (R i=;i<len;++i)
         {
             int c=s[i]-'a';
             if(!ch[x][c])
                 ch[x][c]=++cnt;
             x=ch[x][c];
         }
         ans=(ans+vis[x]*len)%mod;
         vis[x]++;
     }
     void upd (int x,int dep)
     {
         s[x]=vis[x];
         for (R i=;i<;++i)
             if(ch[x][i])
             {
                 upd(ch[x][i],dep+);
                 ans=(ans+1LL*dep*s[x]%mod*(s[ ch[x][i] ]))%mod;
                 s[x]+=s[ ch[x][i] ];
             }
     }
 }T;

 int main()
 {
     scanf("%d%d",&n,&q);
     for (R i=;i<=n;++i)
     {
         scanf("%s",s);
         T.ins(s);
     }
     T.upd(,);
     c[][]=;
     for (R i=;i<=n;++i)
     {
         c[i][]=;
         for (R j=;j<=min(i,);++j)
             c[i][j]=(c[i-][j]+c[i-][j-])%mod;
     }
     for (R i=;i<=q;++i)
     {
         scanf("%d",&k);
         printf("%lld\n",1LL*c[n-][k]*T.ans%mod);
     }
     return ;
 }

D