2018.8.26 练习赛

T1 函数
题面：
题解：
堆code:

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<queue>
 5 #define maxn 100005
 6 #define db long double
 7 using namespace std;
 8 struct func {
 9     db k,v;
10 };
11 bool operator >(func a,func b) {
12     return a.v>b.v;
13 }
14 db nb[maxn];
15 priority_queue<func,vector<func>,greater<func> > q;
16 db l,r,n,x;
17 int main() {
18     scanf("%Lf%Lf%Lf%Lf",&l,&r,&x,&n);
19     for(db i=l; i<=r; i+=1) q.push((func) {
20         i,i*sqrt(x)+1
21     });
22     while(n>1) {
23         func t=q.top();
24         q.pop();
25         q.push((func) {
26             t.k,t.v+1.0
27         });
28         n--;
29     }
30     printf("%.2Lf",q.top().v);
31 }

View Code

T2 情侣
题面：
题解：
具体转移实现见code注释部分：

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<ctype.h>
 4 #define ll long long
 5 #define inv2 499122177
 6 #define mod 998244353
 7 using namespace std;
 8 
 9 ll t;
10 ll D[5005];
11 int f[5005][5005];
12 
13 char buf[1<<20],*p1,*p2;
14 inline char gc() {
15     return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?0:*p1++;
16 }
17 
18 template<typename T>
19 void read(T &x) {
20     char tt;
21     bool flag=0;
22     while(!isdigit(tt=gc())&&tt!='-');
23     tt=='-'?(x=0,flag=1):(x=tt-'0');
24     while(isdigit(tt=gc())) x=x*10+tt-'0';
25     if(flag) x=-x;
26 }
27 
28 ll ksm(ll a,ll b) {
29     ll ans=1;
30     for(; b; b>>=1,a=(long long)a*a%mod)
31         if(b&1) ans=(long long)ans*a%mod;
32     return ans;
33 }
34 //f[i][j]
35 //空2*i+1
36 //情侣空 j
37 //非情侣空2*i+1-j
38 //f[1][1]=1
39 //f[i][j]+=
40 //在一起:
41 //不拆 f[i-1][j-1]*(2*(i-1)+(1-(j-1)))=f[i-1][j-1]*(2*i-j)
42 //拆 f[i-1][j]*j
43 //不在一起:
44 //不拆 f[i-1][j]*(2*(i-1)+1-j)*(2*(i-1)-j)>>1
45 //拆一对  f[i-1][j+1]*(j+1)*(2*(i-1)-j)
46 //拆两对  f[i-1][j+2]*(j+2)*(j+1)>>1
47 void sent() {
48     f[1][1]=1;
49     for(ll i=2; i<=5000; i++)
50         for(ll j=0; j<=i; j++) {
51             if(j)f[i][j]=(f[i][j]+f[i-1][j-1]*(2*i-j)%mod)%mod;
52             f[i][j]=(f[i][j]+f[i-1][j]*j%mod)%mod;
53             f[i][j]=(f[i][j]+f[i-1][j]*(2*(i-1)+1-j)%mod*(2*(i-1)-j)%mod*inv2%mod)%mod;
54             f[i][j]=(f[i][j]+f[i-1][j+1]*(j+1)%mod*(2*(i-1)-j)%mod)%mod;
55             f[i][j]=(f[i][j]+f[i-1][j+2]*(j+2)%mod*(j+1)%mod*inv2%mod);
56         }
57 
58     fact[0]=1;
59     for(int i=1; i<=10000; i++) fact[i]=fact[i-1]*i%mod;
60     ll now=inv2;
61     for(int i=1; i<=5000; i++) {
62         ni[i]=fact[i*2]*now%mod;
63         now=now*inv2%mod;
64     }
65 }
66 
67 int main() {
68     read(t);
69     sent();
70     while(t--) {
71         ll n,d;
72         read(n),read(d);
73         D[0]=1;
74         ll ans=f[n][0];
75         for(ll i=1; i<=n; i++) {
76             D[i]=D[i-1]*d%mod;
77             ans=(ans+D[i]*f[n][i]%mod)%mod;
78         }
79         printf("%lld\n",ans*ksm(ni[n],mod-2)%mod);
80     }
81     return 0;
82 }

View Code

T3 军训（今日毒瘤）
题面：
题解：
std code：

  1 #pragma GCC optimize(3)
  2 #include<stdio.h>
  3 #include<ctype.h>
  4 #include<cmath>
  5 #include<cstring>
  6 #define RG register
  7 #define maxn 300005
  8 #define db double
  9 using namespace std;
 10 char buf[1<<20],*p1,*p2;
 11 #define GC (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
 12 inline int R_int()
 13 {
 14     char t=GC;
 15     int x=0,f=1;
 16     while(!isdigit(t)&&t!='-') t=GC;
 17     if(t=='-') f=-1,t=GC;
 18     while(isdigit(t)) x=x*10+t-48,t=GC;
 19     return f*x;
 20 }
 21 inline char R_alp()
 22 {
 23     char t=GC;
 24     while(!isalpha(t)) t=GC;
 25     return t;
 26 }
 27 struct op{char id;db v1,v2;}a[maxn];
 28 typedef db Mat[4][4];
 29 typedef db Vec[4];
 30 const db Pi=acos(-1.0);
 31 int ls[maxn*2],rs[maxn*2],tot=1,n,m;
 32 op lazy[maxn*2];
 33 int pd[maxn*2];
 34 Mat lval[maxn*2],rval[maxn*2];
 35 inline void Mulmat(Mat &at,Mat &s,Mat &c)
 36 {
 37     static db b[4][4],a[4][4];
 38     for(RG int i=1;i<=3;i++)
 39         for(RG int j=1;j<=3;j++)
 40             b[i][j]=s[j][i],a[i][j]=at[i][j];
 41     for(RG int i=1;i<=3;i++)
 42         for(RG int j=1;j<=3;j++)
 43         {
 44             RG db t=0;
 45             for(db *x=a[i]+1,*y=b[j]+1,k=1;k<=3;++x,++y,++k) t+=*x**y;
 46             c[i][j]=t;
 47         }
 48 }
 49 inline void Mulvec(Vec &a,Mat &s,Vec &c)
 50 {
 51     static db b[4][4];
 52     Vec tmp;
 53     for(RG int i=1;i<=3;i++)
 54         for(RG int j=1;j<=3;j++)
 55             b[i][j]=s[j][i];
 56     for(RG int i=1;i<=3;i++)
 57     {
 58         register db t=0.0;
 59         int k=1;
 60         for(db *x=b[i]+1,*y=a+1;k<=3;++x,++y,++k)
 61             t+=*x**y;
 62         tmp[i]=t;
 63     }
 64     for(RG int i=1;i<=3;i++) c[i]=tmp[i];
 65 } 
 66 inline void get(Mat &a,op b,int t)
 67 {
 68     memset(a,0,sizeof(a));
 69     a[1][1]=a[2][2]=a[3][3]=1;
 70     switch(b.id)
 71     {
 72         case 'R': {a[3][1]=b.v1*(db)t;break;}
 73         case 'L': {a[3][1]=-b.v1*(db)t;break;}
 74         case 'U': {a[3][2]=b.v1*(db)t;break;}
 75         case 'D': {a[3][2]=-b.v1*(db)t;break;}
 76         case 'X': {a[2][2]=((t&1)?-1:1);break;}
 77         case 'Y': {a[1][1]=((t&1)?-1:1);break;}
 78         case 'C':
 79         {
 80             int tmp=(int)floor(b.v1*(db)t);
 81             tmp%=360;
 82             db Rad=1.0*(db)tmp*Pi/180.0;
 83             a[1][1]=cos(Rad),a[1][2]=sin(Rad),a[2][1]=-sin(Rad),a[2][2]=cos(Rad);
 84             break;
 85         }
 86         case 'O': {if(t&1) a[1][1]=-1,a[2][2]=-1,a[3][1]=2*b.v1,a[3][2]=2*b.v2,a[3][3]=1;break;}
 87     }
 88 }
 89 void Read(op &x)
 90 {
 91     char cmd=R_alp();
 92     x.id=cmd;
 93     if(cmd=='U'||cmd=='D'||cmd=='L'||cmd=='R'||cmd=='C') x.v1=(db)R_int();
 94     else if(cmd=='O') x.v1=(db)R_int(),x.v2=(db)R_int();
 95 }
 96 inline void update(int p)
 97 {
 98     Mulmat(lval[ls[p]],lval[rs[p]],lval[p]);
 99     Mulmat(rval[rs[p]],rval[ls[p]],rval[p]);
100 }
101 inline void putdown(int p,int l,int r)
102 {
103     int mid=(l+r)>>1;
104     get(lval[ls[p]],lazy[p],mid-l+1);
105     get(rval[ls[p]],lazy[p],mid-l+1);
106     get(lval[rs[p]],lazy[p],r-mid);
107     get(rval[rs[p]],lazy[p],r-mid);
108     lazy[ls[p]]=lazy[rs[p]]=lazy[p];
109     pd[ls[p]]=1,pd[rs[p]]=1;
110     pd[p]=0;
111 }
112 void build(int p,int l,int r)
113 {
114     if(l==r)
115     {
116         get(lval[p],a[l],1);
117         get(rval[p],a[l],1);
118         return;
119     }
120     int mid=(l+r)>>1;
121     ls[p]=++tot,build(ls[p],l,mid);
122     rs[p]=++tot,build(rs[p],mid+1,r);
123     update(p);
124 }
125 void modify(int p,int l,int r,int x,op s)
126 {
127     if(l==r) {get(lval[p],s,1),get(rval[p],s,1);return;}
128     if(pd[p]) putdown(p,l,r);
129     int mid=(l+r)>>1;
130     if(x<=mid) modify(ls[p],l,mid,x,s);
131     else modify(rs[p],mid+1,r,x,s);
132     update(p);
133 }
134 void mod_range(int p,int l,int r,int x,int y,op s)
135 {
136     if(x<=l&&r<=y)
137     {
138         get(lval[p],s,r-l+1),get(rval[p],s,r-l+1);
139         lazy[p]=s,pd[p]=1;
140         return;
141     }
142     if(pd[p]) putdown(p,l,r);
143     int mid=(l+r)>>1;
144     if(x<=mid&&y>=l) mod_range(ls[p],l,mid,x,y,s);
145     if(y>mid&&x<=r) mod_range(rs[p],mid+1,r,x,y,s);
146     update(p);
147 }
148 Mat ret;
149 void query(int p,int l,int r,int x,int y,int f)
150 {
151     if(x<=l&&r<=y) {Mulmat(ret,f?lval[p]:rval[p],ret);return;}
152     if(pd[p]) putdown(p,l,r);
153     int mid=(l+r)>>1;
154     if(f)
155     {
156         if(x<=mid&&y>=l) query(ls[p],l,mid,x,y,f);
157         if(y>mid&&x<=r) query(rs[p],mid+1,r,x,y,f);        
158     }
159     else
160     {
161         if(y>mid&&x<=r) query(rs[p],mid+1,r,x,y,f);    
162         if(x<=mid&&y>=l) query(ls[p],l,mid,x,y,f);
163     }
164 }
165 void ask(db x,db y,int l,int r,int f)
166 {
167     memset(ret,0,sizeof(ret));
168     ret[1][1]=ret[2][2]=ret[3][3]=1;
169     Vec tmp={0,x,y,1};
170     query(1,1,n,l,r,f);
171     Mulvec(tmp,ret,tmp);
172     printf("%.1lf %.1lf\n",tmp[1],tmp[2]);
173 }
174 int main()
175 {
176     n=R_int(),m=R_int();
177     for(int i=1;i<=n;i++)
178         Read(a[i]);
179     build(1,1,n);
180     for(int i=1;i<=m;i++)
181     {
182         op t;
183         char cmd=R_alp();
184         if(cmd=='M')
185         {
186             int p=R_int();
187             Read(t);
188             modify(1,1,n,p,t);
189         }
190         else if(cmd=='E')
191         {
192             int l=R_int(),r=R_int();
193             Read(t);
194             mod_range(1,1,n,l,r,t);
195         }
196         else if(cmd=='A'||cmd=='B')
197         {
198             db x=(db)R_int(),y=(db)R_int();
199             int l=R_int(),r=R_int();
200             cmd=='A'?ask(x,y,l,r,1):ask(x,y,l,r,0);
201         }
202     }
203     return 0;
204 }

View Code

未ac code（不完整，函数解法）……

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<algorithm>
  4 #include<ctype.h>
  5 #define l(x) x<<1
  6 #define r(x) x<<1|1
  7 #define ll long long
  8 #define ld double
  9 using namespace std;
 10 
 11 char buf[1<<20],*p1,*p2;
 12 inline char gc() {
 13     return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?0:*p1++;
 14 }
 15 
 16 ll n,m;
 17 ll M=1;
 18 ld x,y;
 19 const ld pi=acos(-1);
 20 struct node {
 21     ld A,B,C;
 22     node() {
 23         A=1;
 24         B=0;
 25         C=0;
 26     }
 27 };
 28 
 29 struct point {
 30     node x,y;
 31 } ca[300005<<2],cb[300005<<2];
 32 
 33 template<typename T>
 34 void read(T &x) {
 35     char tt;
 36     bool flag=0;
 37     while(!isdigit(tt=gc())&&tt!='-');
 38     tt=='-'?(x=0,flag=1):(x=tt-'0');
 39     while(isdigit(tt=gc())) x=x*10+tt-'0';
 40     if(flag) x=-x;
 41 }
 42 
 43 void up(point &a,ld t) {
 44     a.y.C+=t;
 45 }
 46 
 47 void down(point &a,ld t) {
 48     a.y.C-=t;
 49 }
 50 
 51 void left(point &a,ld t) {
 52     a.x.C+=t;
 53 }
 54 
 55 void right(point &a,ld t) {
 56     a.x.C-=t;
 57 }
 58 
 59 void dcx(point &a) {
 60     a.x.A=-a.x.A;
 61     a.x.B=-a.x.B;
 62     a.x.C=-a.x.C;
 63 }
 64 
 65 void dcy(point &a) {
 66     a.y.A=-a.y.A;
 67     a.y.B=-a.y.B;
 68     a.y.C=-a.y.C;
 69 }
 70 
 71 void dcxy(point &a,ld d,ld k) {
 72     a.x.A=-a.x.A;
 73     a.x.B=-a.x.B;
 74     a.x.C=2*d-a.x.C;
 75     a.y.A=-a.y.A;
 76     a.y.B=-a.y.B;
 77     a.y.C=2*k-a.x.C;
 78 }
 79 
 80 void turn(point &a,ld c) {
 81     c*=(pi/180);
 82     ld A=a.x.A,B=a.x.B,C=a.x.C,E=a.y.A,F=a.y.B,G=a.y.C;
 83     a.x.A=(A*cos(c)-E*sin(c));
 84     a.x.B=(B*cos(c)-F*sin(c));
 85     a.x.C=(C*cos(c)-G*sin(c));
 86     a.y.A=(A*sin(c)+E*cos(c));
 87     a.y.B=(B*sin(c)+F*cos(c));
 88     a.y.C=(C*sin(c)+G*cos(c));
 89 }
 90 
 91 point merge(point a,point b) {
 92     point tmp;
 93     ld A=b.x.A,B=b.x.B,C=b.x.C,D=b.y.A,E=b.y.A,F=b.y.A;
 94     ld aa=a.x.A,bb=a.x.A,c=a.x.A,d=a.y.A,e=a.y.A,f=a.y.A;
 95     tmp.x.A=(A*aa+B*d);
 96     tmp.x.B=(A*bb+B*e);
 97     tmp.x.C=(C+A*c+B*f);
 98 
 99     tmp.y.A=(D*aa+E*d);
100     tmp.y.B=(D*bb+E*e);
101     tmp.y.C=(F+D*c+E*f);
102     return tmp;
103 }
104 
105 void build() {
106     for(int i=M-1; i>=1; i--) {
107         ca[i]=merge(ca[l(i)],ca[r(i)]);
108         cb[i]=merge(cb[l(i)],cb[r(i)]);
109     }
110 }
111 
112 int main() {
113     scanf("%lld%lld",&n,&m);
114     while(M<=n)M<<=1;
115     for(int i=1; i<=n; i++) {
116         char ch;
117         ld xd,yd;
118         scanf("\n%c",&ch);
119         if(ch=='U') scanf("%lf",&xd),up(ca[M+i],xd),up(cb[M+n-i+1],xd);
120         else if(ch=='D') scanf("%lf",&xd),down(ca[M+i],xd),down(cb[M+n-i+1],xd);
121         else if(ch=='L') scanf("%lf",&xd),left(ca[M+i],xd),left(cb[M+n-i+1],xd);
122         else if(ch=='R') scanf("%lf",&xd),right(ca[M+i],xd),right(cb[M+n-i+1],xd);
123         else if(ch=='O') scanf("%lf%lf",&xd,&yd),dcxy(ca[M+i],xd,yd),dcxy(cb[M+n-i+1],xd,yd);
124         else if(ch=='C') scanf("%lf",&xd),turn(ca[M+i],xd),turn(cb[M+n-i+1],xd);
125         else if(ch=='X') dcx(ca[M+i]),dcx(cb[M+n-i+1]);
126         else if(ch=='Y') dcy(ca[M+i]),dcy(cb[M+n-i+1]);
127     }
128     build();
129     while(m--) {
130         char cc;
131         scanf("\n%c",&cc);
132         if(cc=='M') {
133             int p,xd,yd;
134             char ch;
135             scanf("%d",&p);
136             
137         } else if(cc=='E');
138         else if(cc=='A');
139         else if(cc=='B');
140     }
141 }

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posted @ 2018-08-26 22:24 Katoumegumi 阅读(69) 评论(0) 编辑收藏举报

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2018.8.26 练习赛

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