csps模拟73-74

模拟73：

T1：哔～～～～～～～～～～～～～～～～～～～～

sb模拟，然而一个小细节打炸了，不想解释（吐嘈大样例没有右移）。。。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int a[][],b[][],pd[][];
 int n,m,x,y,z,d,k,v,r;
 long long ans;
 void add()
 {
     r=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(!a[i][j])++r;
     k=+k%r;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(!a[i][j]){--k;if(!k){a[i][j]=v;return;}}
 }
 inline bool work0()
 {
     int ok=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=pd[i][j]=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(a[i][j])
             {
                 b[++b[][j]][j]=a[i][j];
                 if(b[][j]!=i)ok=;
             }
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(b[i][j])
             {
                 a[++a[][j]][j]=b[i][j];
                 if(a[a[][j]][j]==a[a[][j]-][j]&&!pd[a[][j]-][j])
                 {
                     a[a[][j]][j]=;
                     a[--a[][j]][j]<<=;
                     pd[a[][j]][j]=;
                     ans+=a[a[][j]][j];
                     ok=;
                 }
             }
     if(!ok)return ;
     return ;
 }
 inline bool work1()
 {
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=a[n-i+][j];
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=b[i][j];

     if(work0())return ;

     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=a[n-i+][j];
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=b[i][j];
     return ;
 }
 inline bool work2()
 {
     int ok=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=pd[i][j]=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(a[i][j]){
                 b[i][++b[i][]]=a[i][j];
                 if(b[i][]!=j)ok=;
             }
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)
             if(b[i][j])
             {
                 a[i][++a[i][]]=b[i][j];
                 if(a[i][a[i][]]==a[i][a[i][]-]&&!pd[i][a[i][]-])
                 {
                     a[i][a[i][]]=;
                     a[i][--a[i][]]<<=;
                     pd[i][a[i][]]=;
                     ans+=a[i][a[i][]];
                     ok=;
                 }
             }
     if(!ok)return ;
     return ;
 }
 inline bool work3()
 {
     int ok=;
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=a[i][n-j+];
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=b[i][j];

     if(work2())return ;

     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)b[i][j]=a[i][n-j+];
     for( int i=;i<=n;++i)
         for( int j=;j<=n;++j)a[i][j]=b[i][j];
     return ;
 }
 void pr()
 {
     puts("\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n");
     for(int i=;i<=n;puts(""),++i)
         for(int j=;j<=n;++j)printf("%4d",a[i][j]);
     puts("");
 }
 int main()
 {
     srand(time());
     scanf("%d",&n);
     x=rand()%n+;y=rand()%n+;
     a[x][y]=;
     while()
     {
         pr();k=rand()%+;v=;
         char c=getchar();
         while(c!='w'&&c!='a'&&c!='s'&&c!='d')c=getchar();
         switch(c)
         {
             case 'w':{
                 if(work0()){
                     puts("Skyh have aked IOI");
                     return ;
                 }
                 add();
                 break;
             }
             case 's':{
                 if(work1()){
                     puts("Skyh have aked IOI");
                     return ;
                 }
                 add();
                 break;
             }
             case 'a':{
                 if(work2()){
                     puts("Skyh have aked IOI");
                     return ;
                 }
                 add();
                 break;
             }
             case 'd':{
                 if(work3()){
                     puts("Skyh have aked IOI");
                     return ;
                 }
                 add();
                 break;
             }
         }
     }
     return ;
 }

2048

T2：结论题，dp，数据结构优化，秒切

 //反正法证明段数不会大于2。。。
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define ll long long
 #define N 100050
 using namespace std;
 int n;
 ll ct[N],f[N],dp[N],ans;
 int lsh[N],ls,a[N];
 void init()
 {
     sort(lsh+,lsh+n+);
     ls=unique(lsh+,lsh+n+)-lsh-;
     for(int i=;i<=n;++i)
         a[i]=lower_bound(lsh+,lsh+ls+,a[i])-lsh;
 }
 inline int add(int x,long long v)
 {
     while(x<=n)
     {
         ct[x]=max(ct[x],v);
         x+=x&-x;
     }
 }
 inline ll ask(int x)
 {
     ll ret=;
     while(x)
     {
         if(ct[x]>ret)ret=ct[x];
         x-=x&-x;
     }
     return ret;
 }
 int main()
 {
     scanf("%d",&n);
     for(register int i=;i<=n;++i)
     {
         scanf("%d",&a[i]);
         lsh[++ls]=a[i];
     }
     init();
     for(register int i=;i<=n;++i)
     {
         f[i]=ask(a[i]-)+lsh[a[i]];
         add(a[i],f[i]);
         if(f[i-]>f[i])f[i]=f[i-];
     }
     for(register int i=;i<=n;++i)ct[i]=;
     for(register int i=n;i;--i)
     {
         dp[i]=ask(a[i]-)+lsh[a[i]];add(a[i],dp[i]);
         if(dp[i+]>dp[i])dp[i]=dp[i+];
     }
     for(register int i=;i<=n;++i)
     {
         ans=max(ans,f[i]+max(f[i],dp[i+]));
     }
     double t=0.5*ans;
     printf("%.3lf\n",t);
     return ;
 }

T3：神题，对高考数学涉及较多，主要考察三角函数应用。暴力枚举弧度数即可A掉（数据较水，莫要吐嘈细节）

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<cstdlib>
 #define N 55
 using namespace std;
 const int inf=;
 int n,m,fa[N];
 struct node{
     int x,y;
     double a,b;
     double val,z,ta,ct,si,co;
     inline void init()
     {
         z=sqrt(1.0*a*a+1.0*b*b);
         ta=1.0*b/a;ct=1.0*a/b;
         si=1.0*b/z;co=1.0*a/z;
     }
     friend bool operator < (const node &c,const node &d){return c.val<d.val;}
 }q[],t;
 double ans;
 int getfa(int x)
 {
     if(fa[x]==x)return x;
     return fa[x]=getfa(fa[x]);
 }
 int main()
 {
     srand(time());
     scanf("%d%d",&n,&m);int ok=;
     for(int i=;i<=m;++i)
         scanf("%d%d%lf%lf",&q[i].x,&q[i].y,&q[i].a,&q[i].b),q[i].init();
     random_shuffle(q+,q+m+);
     double a,b;
     for(double i=;i<=;i+=0.7)
     {
         a=b=0.0;
         t.z=1000.0;t.a=t.z*cos(i);t.b=t.z*sin(i);
         t.init();
         for(int j=;j<=m;++j)
             q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si);
         for(int j=;j<=n;++j)fa[j]=j;
         sort(q+,q+m+);
         for(int j=;j<=m;++j)
         {
             if(getfa(q[j].x)==getfa(q[j].y))continue;
             fa[getfa(q[j].x)]=getfa(q[j].y);
             a+=q[j].a;b+=q[j].b;
         }
         ans=max(ans,sqrt(1ll*a*a+1ll*b*b));
     }
     printf("%.6lf\n",ans);
 }

正解skyh说挺简单的不用解释。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<cstdlib>
 #include<queue>
 #define N 55
 using namespace std;
 const int inf=;
 int n,m,fa[N];
 struct node{
     int x,y;
     double a,b;
     double val,z,ta,ct,si,co;
     inline void init()
     {
         z=sqrt(1.0*a*a+1.0*b*b);
         ta=1.0*b/a;ct=1.0*a/b;
         si=1.0*b/z;co=1.0*a/z;
     }
     friend bool operator < (const node &c,const node &d){return c.val<d.val;}
 }q[],t;
 double ans;
 int getfa(int x)
 {
     if(fa[x]==x)return x;
     return fa[x]=getfa(fa[x]);
 }
 inline double Gauss(double x,double y,double xx,double yy)
 {
     double k,b;
     b=yy-y*(xx/x);
     b*=x/xx;
     k=(y-b)/x;
     return k;
 }
 queue<double>qq;
 void work1()
 {
     double a,b;
     for(double i=;i<=;i+=0.7)
     {
         a=b=0.0;
         t.z=1000.0;t.a=t.z*cos(i);t.b=t.z*sin(i);
         t.init();
         for(int j=;j<=m;++j)
             q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si);
         for(int j=;j<=n;++j)fa[j]=j;
         sort(q+,q+m+);
         for(int j=;j<=m;++j)
         {
             if(getfa(q[j].x)==getfa(q[j].y))continue;
             fa[getfa(q[j].x)]=getfa(q[j].y);
             a+=q[j].a;b+=q[j].b;
         }
         ans=max(ans,sqrt(1ll*a*a+1ll*b*b));
     }
     printf("%.6lf\n",ans);
 }
 int main()
 {
     srand(time());
     scanf("%d%d",&n,&m);int ok=;
     for(int i=;i<=m;++i)
         {scanf("%d%d%lf%lf",&q[i].x,&q[i].y,&q[i].a,&q[i].b),q[i].init();if(q[i].b)ok=;}
     double a,b,tt;
     if(ok){work1();return ;}
     for(int i=;i<=m;++i)
     {
         for(int j=;j<=m;++j)
         {
             if(j==i||q[i].b==q[j].b)continue;
             tt=Gauss(q[i].a,q[i].b,q[j].a,q[j].b);
             tt=-1.0/tt;
             qq.push(tt);
         }
     }
     while(!qq.empty())
     {
         a=b=0.0;t.a=1.0;t.b=t.a*(qq.front()+1e-);
         t.init();
         for(int j=;j<=m;++j)
             q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si);
         for(int j=;j<=n;++j)fa[j]=j;
         sort(q+,q+m+);
         for(int j=;j<=m;++j)
         {
             if(getfa(q[j].x)==getfa(q[j].y))continue;
             fa[getfa(q[j].x)]=getfa(q[j].y);
             a+=q[j].a;b+=q[j].b;
         }
         ans=max(ans,sqrt(1ll*a*a+1ll*b*b));
         a=b=0.0;t.a=-1.0;t.b=t.a*(qq.front()+1e-);qq.pop();
         t.init();
         for(int j=;j<=m;++j)
             q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si);
         for(int j=;j<=n;++j)fa[j]=j;
         sort(q+,q+m+);
         for(int j=;j<=m;++j)
         {
             if(getfa(q[j].x)==getfa(q[j].y))continue;
             fa[getfa(q[j].x)]=getfa(q[j].y);
             a+=q[j].a;b+=q[j].b;
         }
         ans=max(ans,sqrt(1ll*a*a+1ll*b*b));
     }
     printf("%.6lf\n",ans);
 }

正解

其实还可以，不太恶心

模拟74：

（这里用来写脏话）

T1&T3:水题。。。

T2：神题

时间分配：T1+T3=1h，T2=2.5h

得分分布：T1+T3=200pts，T2=30pts

T2：发现玩具构成一棵树，求树高的期望。

比较难想的其实是把新加入的点当作根节点，而不是接到其他节点之后

题解：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #define N 250
 using namespace std;
 inline int read()
 {
     char c=getchar();int s=;
     while(c>''||c<'')c=getchar();
     while(c>=''&&c<='')s=s*+c-'',c=getchar();
     return s;
 }
 int n=read();
 const int mod=read();
 int ans;
 inline int qpow(int d,int z)
 {
     int ret=;
     for(;z;z>>=,d=1ll*d*d%mod)
         if(z&)ret=1ll*ret*d%mod;
     return ret;
 }
 int inv[N];
 int alv;
 int dp[N][N];
 int f[N][N];
 int g[N][N];
 int main()
 {
     inv[]=;
     for(int i=;i<=n;++i)inv[i]=qpow(i,mod-);
     alv=;
     for(int i=;i<n;++i)alv=1ll*alv*inv[i]%mod;
     dp[][]=;
     for(int i=;i<=n;++i){
         for(int j=;j<=i;++j){
             dp[i][j]=1ll*inv[i]*(1ll*dp[i-][j-]*(j-)%mod+1ll*dp[i-][j]*(i-j)%mod)%mod;
             if(dp[i][j]>=mod)dp[i][j]-=mod;
         }
     }
     for(int i=;i<=n;++i)g[][i]=f[][i]=g[][i]=;
     for(int i=;i<=n;++i){
         for(int j=;j<n;++j){f[i][j]=g[i-][j-];}
         for(int j=;j<n;++j){
             for(int k=;k<=i;++k){
                 g[i][j]+=1ll*g[i-k][j]*f[k][j]%mod*dp[i][k]%mod;
                 if(g[i][j]>=mod)g[i][j]-=mod;
             }
         }
     }
     for(int i=;i<n;++i)
     {
         ans+=1ll*(f[n][i]-f[n][i-])*i%mod;
         if(ans>=mod)ans-=mod;
         if(ans<)ans+=mod;
     }
     cout<<ans<<endl;
     return ;
 }

发现题解中给的dp数组dp[i][j]=inv[i];证明好像可以用数学归纳法。