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）。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int a[10][10],b[10][10],pd[10][10]; 7 int n,m,x,y,z,d,k,v,r; 8 long long ans; 9 void add() 10 { 11 r=0; 12 for( int i=1;i<=n;++i) 13 for( int j=1;j<=n;++j) 14 if(!a[i][j])++r; 15 k=1+k%r; 16 for( int i=1;i<=n;++i) 17 for( int j=1;j<=n;++j) 18 if(!a[i][j]){--k;if(!k){a[i][j]=v;return;}} 19 } 20 inline bool work0() 21 { 22 int ok=0; 23 for( int i=0;i<=n;++i) 24 for( int j=0;j<=n;++j)b[i][j]=pd[i][j]=0; 25 for( int i=1;i<=n;++i) 26 for( int j=1;j<=n;++j) 27 if(a[i][j]) 28 { 29 b[++b[0][j]][j]=a[i][j]; 30 if(b[0][j]!=i)ok=1; 31 } 32 for( int i=0;i<=n;++i) 33 for( int j=0;j<=n;++j)a[i][j]=0; 34 for( int i=1;i<=n;++i) 35 for( int j=1;j<=n;++j) 36 if(b[i][j]) 37 { 38 a[++a[0][j]][j]=b[i][j]; 39 if(a[a[0][j]][j]==a[a[0][j]-1][j]&&!pd[a[0][j]-1][j]) 40 { 41 a[a[0][j]][j]=0; 42 a[--a[0][j]][j]<<=1; 43 pd[a[0][j]][j]=1; 44 ans+=a[a[0][j]][j]; 45 ok=1; 46 } 47 } 48 if(!ok)return 1; 49 return 0; 50 } 51 inline bool work1() 52 { 53 for( int i=0;i<=n;++i) 54 for( int j=0;j<=n;++j)b[i][j]=a[n-i+1][j]; 55 for( int i=0;i<=n;++i) 56 for( int j=0;j<=n;++j)a[i][j]=b[i][j]; 57 58 if(work0())return 1; 59 60 for( int i=0;i<=n;++i) 61 for( int j=0;j<=n;++j)b[i][j]=a[n-i+1][j]; 62 for( int i=0;i<=n;++i) 63 for( int j=0;j<=n;++j)a[i][j]=b[i][j]; 64 return 0; 65 } 66 inline bool work2() 67 { 68 int ok=0; 69 for( int i=0;i<=n;++i) 70 for( int j=0;j<=n;++j)b[i][j]=pd[i][j]=0; 71 for( int i=1;i<=n;++i) 72 for( int j=1;j<=n;++j) 73 if(a[i][j]){ 74 b[i][++b[i][0]]=a[i][j]; 75 if(b[i][0]!=j)ok=1; 76 } 77 for( int i=0;i<=n;++i) 78 for( int j=0;j<=n;++j)a[i][j]=0; 79 for( int i=1;i<=n;++i) 80 for( int j=1;j<=n;++j) 81 if(b[i][j]) 82 { 83 a[i][++a[i][0]]=b[i][j]; 84 if(a[i][a[i][0]]==a[i][a[i][0]-1]&&!pd[i][a[i][0]-1]) 85 { 86 a[i][a[i][0]]=0; 87 a[i][--a[i][0]]<<=1; 88 pd[i][a[i][0]]=1; 89 ans+=a[i][a[i][0]]; 90 ok=1; 91 } 92 } 93 if(!ok)return 1; 94 return 0; 95 } 96 inline bool work3() 97 { 98 int ok=0; 99 for( int i=0;i<=n;++i) 100 for( int j=0;j<=n;++j)b[i][j]=a[i][n-j+1]; 101 for( int i=0;i<=n;++i) 102 for( int j=0;j<=n;++j)a[i][j]=b[i][j]; 103 104 if(work2())return 1; 105 106 for( int i=0;i<=n;++i) 107 for( int j=0;j<=n;++j)b[i][j]=a[i][n-j+1]; 108 for( int i=0;i<=n;++i) 109 for( int j=0;j<=n;++j)a[i][j]=b[i][j]; 110 return 0; 111 } 112 void pr() 113 { 114 puts(" "); 115 for(int i=1;i<=n;puts(""),++i) 116 for(int j=1;j<=n;++j)printf("%4d",a[i][j]); 117 puts(""); 118 } 119 int main() 120 { 121 srand(time(0)); 122 scanf("%d",&n); 123 x=rand()%n+1;y=rand()%n+1; 124 a[x][y]=2; 125 while(1) 126 { 127 pr();k=rand()%233+1;v=2; 128 char c=getchar(); 129 while(c!='w'&&c!='a'&&c!='s'&&c!='d')c=getchar(); 130 switch(c) 131 { 132 case 'w':{ 133 if(work0()){ 134 puts("Skyh have aked IOI"); 135 return 0; 136 } 137 add(); 138 break; 139 } 140 case 's':{ 141 if(work1()){ 142 puts("Skyh have aked IOI"); 143 return 0; 144 } 145 add(); 146 break; 147 } 148 case 'a':{ 149 if(work2()){ 150 puts("Skyh have aked IOI"); 151 return 0; 152 } 153 add(); 154 break; 155 } 156 case 'd':{ 157 if(work3()){ 158 puts("Skyh have aked IOI"); 159 return 0; 160 } 161 add(); 162 break; 163 } 164 } 165 } 166 return 0; 167 }
T2：结论题，dp，数据结构优化，秒切
1 //反正法证明段数不会大于2。。。 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 #define ll long long 7 #define N 100050 8 using namespace std; 9 int n; 10 ll ct[N],f[N],dp[N],ans; 11 int lsh[N],ls,a[N]; 12 void init() 13 { 14 sort(lsh+1,lsh+n+1); 15 ls=unique(lsh+1,lsh+n+1)-lsh-1; 16 for(int i=1;i<=n;++i) 17 a[i]=lower_bound(lsh+1,lsh+ls+1,a[i])-lsh; 18 } 19 inline int add(int x,long long v) 20 { 21 while(x<=n) 22 { 23 ct[x]=max(ct[x],v); 24 x+=x&-x; 25 } 26 } 27 inline ll ask(int x) 28 { 29 ll ret=0; 30 while(x) 31 { 32 if(ct[x]>ret)ret=ct[x]; 33 x-=x&-x; 34 } 35 return ret; 36 } 37 int main() 38 { 39 scanf("%d",&n); 40 for(register int i=1;i<=n;++i) 41 { 42 scanf("%d",&a[i]); 43 lsh[++ls]=a[i]; 44 } 45 init(); 46 for(register int i=1;i<=n;++i) 47 { 48 f[i]=ask(a[i]-1)+lsh[a[i]]; 49 add(a[i],f[i]); 50 if(f[i-1]>f[i])f[i]=f[i-1]; 51 } 52 for(register int i=1;i<=n;++i)ct[i]=0; 53 for(register int i=n;i;--i) 54 { 55 dp[i]=ask(a[i]-1)+lsh[a[i]];add(a[i],dp[i]); 56 if(dp[i+1]>dp[i])dp[i]=dp[i+1]; 57 } 58 for(register int i=1;i<=n;++i) 59 { 60 ans=max(ans,f[i]+max(f[i],dp[i+1])); 61 } 62 double t=0.5*ans; 63 printf("%.3lf ",t); 64 return 0; 65 }
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）
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<cstdlib> 7 #define N 55 8 using namespace std; 9 const int inf=100000; 10 int n,m,fa[N]; 11 struct node{ 12 int x,y; 13 double a,b; 14 double val,z,ta,ct,si,co; 15 inline void init() 16 { 17 z=sqrt(1.0*a*a+1.0*b*b); 18 ta=1.0*b/a;ct=1.0*a/b; 19 si=1.0*b/z;co=1.0*a/z; 20 } 21 friend bool operator < (const node &c,const node &d){return c.val<d.val;} 22 }q[5555],t; 23 double ans; 24 int getfa(int x) 25 { 26 if(fa[x]==x)return x; 27 return fa[x]=getfa(fa[x]); 28 } 29 int main() 30 { 31 srand(time(0)); 32 scanf("%d%d",&n,&m);int ok=1; 33 for(int i=1;i<=m;++i) 34 scanf("%d%d%lf%lf",&q[i].x,&q[i].y,&q[i].a,&q[i].b),q[i].init(); 35 random_shuffle(q+1,q+m+1); 36 double a,b; 37 for(double i=1;i<=360;i+=0.7) 38 { 39 a=b=0.0; 40 t.z=1000.0;t.a=t.z*cos(i);t.b=t.z*sin(i); 41 t.init(); 42 for(int j=1;j<=m;++j) 43 q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si); 44 for(int j=1;j<=n;++j)fa[j]=j; 45 sort(q+1,q+m+1); 46 for(int j=1;j<=m;++j) 47 { 48 if(getfa(q[j].x)==getfa(q[j].y))continue; 49 fa[getfa(q[j].x)]=getfa(q[j].y); 50 a+=q[j].a;b+=q[j].b; 51 } 52 ans=max(ans,sqrt(1ll*a*a+1ll*b*b)); 53 } 54 printf("%.6lf ",ans); 55 }
正解skyh说挺简单的不用解释。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<cstdlib> 7 #include<queue> 8 #define N 55 9 using namespace std; 10 const int inf=100000; 11 int n,m,fa[N]; 12 struct node{ 13 int x,y; 14 double a,b; 15 double val,z,ta,ct,si,co; 16 inline void init() 17 { 18 z=sqrt(1.0*a*a+1.0*b*b); 19 ta=1.0*b/a;ct=1.0*a/b; 20 si=1.0*b/z;co=1.0*a/z; 21 } 22 friend bool operator < (const node &c,const node &d){return c.val<d.val;} 23 }q[5555],t; 24 double ans; 25 int getfa(int x) 26 { 27 if(fa[x]==x)return x; 28 return fa[x]=getfa(fa[x]); 29 } 30 inline double Gauss(double x,double y,double xx,double yy) 31 { 32 double k,b; 33 b=yy-y*(xx/x); 34 b*=x/xx; 35 k=(y-b)/x; 36 return k; 37 } 38 queue<double>qq; 39 void work1() 40 { 41 double a,b; 42 for(double i=1;i<=360;i+=0.7) 43 { 44 a=b=0.0; 45 t.z=1000.0;t.a=t.z*cos(i);t.b=t.z*sin(i); 46 t.init(); 47 for(int j=1;j<=m;++j) 48 q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si); 49 for(int j=1;j<=n;++j)fa[j]=j; 50 sort(q+1,q+m+1); 51 for(int j=1;j<=m;++j) 52 { 53 if(getfa(q[j].x)==getfa(q[j].y))continue; 54 fa[getfa(q[j].x)]=getfa(q[j].y); 55 a+=q[j].a;b+=q[j].b; 56 } 57 ans=max(ans,sqrt(1ll*a*a+1ll*b*b)); 58 } 59 printf("%.6lf ",ans); 60 } 61 int main() 62 { 63 srand(time(0)); 64 scanf("%d%d",&n,&m);int ok=1; 65 for(int i=1;i<=m;++i) 66 {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=0;} 67 double a,b,tt; 68 if(ok){work1();return 0;} 69 for(int i=1;i<=m;++i) 70 { 71 for(int j=1;j<=m;++j) 72 { 73 if(j==i||q[i].b==q[j].b)continue; 74 tt=Gauss(q[i].a,q[i].b,q[j].a,q[j].b); 75 tt=-1.0/tt; 76 qq.push(tt); 77 } 78 } 79 while(!qq.empty()) 80 { 81 a=b=0.0;t.a=1.0;t.b=t.a*(qq.front()+1e-3); 82 t.init(); 83 for(int j=1;j<=m;++j) 84 q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si); 85 for(int j=1;j<=n;++j)fa[j]=j; 86 sort(q+1,q+m+1); 87 for(int j=1;j<=m;++j) 88 { 89 if(getfa(q[j].x)==getfa(q[j].y))continue; 90 fa[getfa(q[j].x)]=getfa(q[j].y); 91 a+=q[j].a;b+=q[j].b; 92 } 93 ans=max(ans,sqrt(1ll*a*a+1ll*b*b)); 94 a=b=0.0;t.a=-1.0;t.b=t.a*(qq.front()+1e-3);qq.pop(); 95 t.init(); 96 for(int j=1;j<=m;++j) 97 q[j].val=q[j].z*(t.co*q[j].co+t.si*q[j].si); 98 for(int j=1;j<=n;++j)fa[j]=j; 99 sort(q+1,q+m+1); 100 for(int j=1;j<=m;++j) 101 { 102 if(getfa(q[j].x)==getfa(q[j].y))continue; 103 fa[getfa(q[j].x)]=getfa(q[j].y); 104 a+=q[j].a;b+=q[j].b; 105 } 106 ans=max(ans,sqrt(1ll*a*a+1ll*b*b)); 107 } 108 printf("%.6lf ",ans); 109 }
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：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #define N 250 7 using namespace std; 8 inline int read() 9 { 10 char c=getchar();int s=0; 11 while(c>'9'||c<'0')c=getchar(); 12 while(c>='0'&&c<='9')s=s*10+c-'0',c=getchar(); 13 return s; 14 } 15 int n=read(); 16 const int mod=read(); 17 int ans; 18 inline int qpow(int d,int z) 19 { 20 int ret=1; 21 for(;z;z>>=1,d=1ll*d*d%mod) 22 if(z&1)ret=1ll*ret*d%mod; 23 return ret; 24 } 25 int inv[N]; 26 int alv; 27 int dp[N][N]; 28 int f[N][N]; 29 int g[N][N]; 30 int main() 31 { 32 inv[0]=1; 33 for(int i=1;i<=n;++i)inv[i]=qpow(i,mod-2); 34 alv=1; 35 for(int i=1;i<n;++i)alv=1ll*alv*inv[i]%mod; 36 dp[1][1]=1; 37 for(int i=2;i<=n;++i){ 38 for(int j=1;j<=i;++j){ 39 dp[i][j]=1ll*inv[i]*(1ll*dp[i-1][j-1]*(j-1)%mod+1ll*dp[i-1][j]*(i-j)%mod)%mod; 40 if(dp[i][j]>=mod)dp[i][j]-=mod; 41 } 42 } 43 for(int i=0;i<=n;++i)g[1][i]=f[1][i]=g[0][i]=1; 44 for(int i=2;i<=n;++i){ 45 for(int j=1;j<n;++j){f[i][j]=g[i-1][j-1];} 46 for(int j=0;j<n;++j){ 47 for(int k=1;k<=i;++k){ 48 g[i][j]+=1ll*g[i-k][j]*f[k][j]%mod*dp[i][k]%mod; 49 if(g[i][j]>=mod)g[i][j]-=mod; 50 } 51 } 52 } 53 for(int i=1;i<n;++i) 54 { 55 ans+=1ll*(f[n][i]-f[n][i-1])*i%mod; 56 if(ans>=mod)ans-=mod; 57 if(ans<0)ans+=mod; 58 } 59 cout<<ans<<endl; 60 return 0; 61 }
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。