CH Round #48 - Streaming #3 (NOIP模拟赛Day1)

A.数三角形

题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/数三角形

题解：暴力枚举三元组，判断是否共线即可，用叉积

代码：

 type cp=record
      x,y:double;
      end;
 var a:array[..] of cp;
     i,j,k,n,ans:longint;
 function cross(a,b,c:cp):double;
  begin
    cross:=(b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
  end;
 procedure init;
  begin
    readln(n);
    for i:= to n do readln(a[i].x,a[i].y);
  end;
 procedure main;
  begin
    ans:=;
    for i:= to n do
     for j:=i+ to n do
      for k:=j+ to n do
       begin
       if cross(a[i],a[j],a[k])<> then inc(ans);
       end;
    writeln(ans);
  end;

 begin
   init;
   main;
 end.                         

B.4和7

题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/4和7

题解：这类似于vijosp1002过河，那题是当距离超过105时缩到105，但我取20也过了，现在也不知道为什么要压。。。

这一道题比较好做，因为只有4和7两个数，由裴蜀定理我们知道 （4，7）=1 所以仅有4和7是可以组合出所有的整数的，

但这题必须保证4和7的系数都必须是非负的。

所以我们从前面开始尝试，看看能发现什么规律。

发现前面有的可以4和7组合出来 比如15，但有的就组合不出来，如13

进一步我们又发现当x>=18时，就都可以组合出来了。我们详细的证明一下：

i） x mod 4=0 直接全用4即可

ii)  x mod 4=1 我们可以用3个7，剩下的全用4，因为这样的数最小是21

iii）x mod 4=2 我们用2个7，剩下的全用4

iiii）x mod 4=3 我们用1个7，剩下的全用4

这样就可以了当相邻两个有药的位置距离相隔≥18时就将它看作18，这样是没有问题的。

考场上没有处理刚开始的边界情况，WA了30分

代码：

 var v,a,b,c,f:array[-..] of int64;
     i,n:longint;
     ans,m,tmp:int64;
 function max(x,y:int64):int64;
  begin
  if x>y then exit(x) else exit(y);
  end;
 procedure sort(l,r:longint);
  var i,j,x,y:longint;
  begin
  i:=l;j:=r;x:=b[(i+j)>>];
  repeat
   while b[i]<x do inc(i);
   while b[j]>x do dec(j);
   if i<=j then
    begin
    y:=a[i];a[i]:=a[j];a[j]:=y;
    y:=b[i];b[i]:=b[j];b[j]:=y;
    inc(i);dec(j);
    end;
  until i>j;
  if i<r then sort(i,r);
  if j>l then sort(l,j);
  end;
 procedure init;
  begin
   readln(n,m);
   for i:= to n do readln(a[i],b[i]);
   sort(,n);
  end;
 procedure main;
  begin
   b[]:=;c[]:=;
   for i:= to n do
    if b[i]-b[i-]>= then c[i]:=c[i-]+
    else c[i]:=c[i-]+b[i]-b[i-];
   fillchar(v,sizeof(v),);
   for i:= to n do inc(v[c[i]],a[i]);
   f[]:=;
   for i:= to c[n] do
    begin
    tmp:=max(f[i-],f[i-]);
    if tmp= then continue;
    f[i]:=tmp+v[i];
    end;
   ans:=;
   for i:= to c[n] do ans:=max(ans,f[i]);
   writeln(ans-);
  end;

 begin
   init;
   main;
 end.     

C.反射镜

题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/反射镜

题解：考场上没想到什么好的算法，于是打了200行的暴力。。。幸亏没写跪，骗到60分，结果标程竟然那么短。。。

代码：

1.暴力

 type cp=record
      x,y,id,idx:longint;
      z:char;
      end;
 var a,b:array[..] of cp;
     now,dir,n,m,i,x,y,tmp:longint;
     c:array[..] of longint;
     dist,t:int64;
     ans:cp;

 function cmp1(a,b:cp):boolean;
  begin
  if a.x=b.x then exit(a.y<b.y) else exit(a.x<b.x);
  end;
 function cmp2(a,b:cp):boolean;
  begin
  if a.y=b.y then exit(a.x<b.x) else exit(a.y<b.y);
  end;
 procedure sort1(l,r:longint);
  var i,j:longint;x,y:cp;
  begin
  i:=l;j:=r;x:=a[(i+j)>>];
  repeat
   while cmp1(a[i],x) do inc(i);
   while cmp1(x,a[j]) do dec(j);
   if i<=j then
    begin
    y:=a[i];a[i]:=a[j];a[j]:=y;
    inc(i);dec(j);
    end;
  until i>j;
  if i<r then sort1(i,r);
  if j>l then sort1(l,j);
  end;
 procedure sort2(l,r:longint);
  var i,j:longint;x,y:cp;
  begin
  i:=l;j:=r;x:=b[(i+j)>>];
  repeat
   while cmp2(b[i],x) do inc(i);
   while cmp2(x,b[j]) do dec(j);
   if i<=j then
    begin
    y:=b[i];b[i]:=b[j];b[j]:=y;
    inc(i);dec(j);
    end;
  until i>j;
  if i<r then sort2(i,r);
  if j>l then sort2(l,j);
  end;
 procedure init;
  begin
  readln(n,m,t);
  for i:= to n do begin readln(a[i].x,a[i].y,a[i].z,a[i].z);a[i].id:=i;end;
  for i:= to n do b[i]:=a[i];
  sort1(,n);
  sort2(,n);
  for i:= to n do c[a[i].id]:=i;
  for i:= to n do b[i].idx:=c[b[i].id];
  for i:= to n do c[b[i].id]:=i;
  for i:= to n do a[i].idx:=c[a[i].id];
  a[n+].x:=maxlongint;a[n+].y:=maxlongint;
  a[].x:=maxlongint;a[].y:=maxlongint;
  b[n+]:=a[n+];b[]:=a[];
  end;
 procedure main;
  begin
   x:=;y:=;
   for i:= to n do if (b[i].y=) and (b[i].x>) then begin now:=i;break;end;
   if now= then
      begin
      writeln(t,' ',);exit;
      end;
   dir:=;dist:=b[now].x;ans:=b[now];
   while true do
    begin
     if dir= then
      begin
       if b[now].z='/' then
         begin
         dir:=;
         now:=b[now].idx;
         tmp:=now+;
         if a[tmp].x<>a[now].x then break;
         if dist+abs(a[tmp].y-a[now].y)>t then break
         else inc(dist,abs(a[tmp].y-a[now].y));
         now:=tmp;ans:=a[now];
         end
       else
         begin
          dir:=;
          now:=b[now].idx;
          tmp:=now-;
          if a[tmp].x<>a[tmp].x then break;
          if dist+abs(a[tmp].y-a[now].y)>t then break
          else inc(dist,abs(a[tmp].y-a[now].y));
          now:=tmp;ans:=a[now];
         end;
      end;
     if dir= then
       begin
        if a[now].z='/' then
          begin
           dir:=;
           now:=a[now].idx;
           tmp:=now+;
           if b[tmp].y<>b[now].y then break;
           if dist+abs(b[tmp].x-b[now].x)>t then break
           else inc(dist,abs(b[tmp].x-b[now].x));
           now:=tmp;ans:=b[now];
          end
        else
          begin
           dir:=;
           now:=a[now].idx;
           tmp:=now-;
           if b[tmp].y<>b[now].y then break;
           if dist+abs(b[tmp].x-b[now].x)>t then break
           else inc(dist,abs(b[tmp].x-b[now].x));
           now:=tmp;ans:=b[now];
          end;
       end;
     if dir= then
        begin
         if b[now].z='/' then
           begin
            dir:=;
            now:=b[now].idx;
            tmp:=now-;
            if a[tmp].x<>a[now].x then break;
            if dist+abs(a[tmp].y-a[now].y)>t then break
            else inc(dist,abs(a[tmp].y-a[now].y));
            now:=tmp;ans:=a[now];
           end
         else
           begin
            dir:=;
            now:=b[now].idx;
            tmp:=now+;
            if a[tmp].x<>a[now].x then break;
            if dist+abs(a[tmp].y-a[now].y)>t then break
            else inc(dist,abs(a[tmp].y-a[now].y));
            now:=tmp;ans:=a[now];
           end;
        end;
     if dir= then
       begin
        if a[now].z='/' then
          begin
           dir:=;
           now:=a[now].idx;
           tmp:=now-;
           if b[tmp].y<>b[now].y then break;
           if dist+abs(b[tmp].x-b[now].x)>t then break
           else inc(dist,abs(b[tmp].x-b[now].x));
           now:=tmp;ans:=b[now];
          end
        else
          begin
           dir:=;
           now:=a[now].idx;
           tmp:=now+;
           if b[tmp].y<>b[now].y then break;
           if dist+abs(b[tmp].x-b[now].x)>t then break
            else inc(dist,abs(b[tmp].x-b[now].x));
           now:=tmp;ans:=b[now];
          end;
       end;
    end;
  if dir= then writeln(ans.x-(t-dist),' ',ans.y);
  if dir= then writeln(ans.x,' ',ans.y+(t-dist));
  if dir= then writeln(ans.x+(t-dist),' ',ans.y);
  if dir= then writeln(ans.x,' ',ans.y-(t-dist));
  end;
 procedure spj;
  begin
  if a[].y<> then writeln(t,' ',)
  else if a[].x< then writeln(t,' ',)
  else if a[].z='/' then writeln(a[].x,' ',t-a[].x)
  else writeln(a[].x,' ',a[].x-t);
  end;

 begin
   init;
   if n= then spj else main;
 end.                            

2.正解

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define UU 0
 #define LL 1
 #define DD 2
 #define RR 3
 //LeftTurn = 1
 //RightTurn = 3
 const int dx[] = {, -, , };
 const int dy[] = {, , -, };
 #define f(x, y, z) for(int x = (y); x <= (z); ++x)
 #define f2(x, y, z) for(int x = (y), asdf = (z); x <= asdf; ++x)
 using std::min;
 using std::abs;

 int N; long long T;
 struct mir{int x, y; long long dr;} xm[], ym[];
 inline bool byx(const mir &a, const mir &b){
     return a.x < b.x || (a.x == b.x && a.y < b.y);
 }
 inline bool byy(const mir &a, const mir &b){
     return a.y < b.y || (a.y == b.y && a.x < b.x);
 }
 inline bool operator ==(const mir &a, const mir &b){
     return a.x == b.x && a.y == b.y;
 }

 inline mir *findxless(const mir a){
     int l = , r = N + ;
     while(l < r){
         int m = ((l + r) >> ) + ;
         if(xm[m] == a) return xm + m;
         else if(byx(xm[m], a)) l = m; else r = m - ;
     }
     if(l <  || l > N || xm[l].x != a.x) return NULL;
     return xm + l;
 }
 inline mir *findxmore(const mir a){
     int l = , r = N + ;
     while(l < r){
         int m = (l + r) >> ;
         if(xm[m] == a) return xm + m;
         else if(byx(a, xm[m])) r = m; else l = m + ;
     }
     if(l <  || l > N || xm[l].x != a.x) return NULL;
     return xm + l;
 }
 inline mir *findyless(const mir a){
     int l = , r = N + ;
     while(l < r){
         int m = ((l + r) >> ) + ;
         if(ym[m] == a) return ym + m;
         else if(byy(ym[m], a)) l = m; else r = m - ;
     }
     if(l <  || l > N || ym[l].y != a.y) return NULL;
     return ym + l;
 }
 inline mir *findymore(const mir a){
     int l = , r = N + ;
     while(l < r){
         int m = (l + r) >> ;
         if(ym[m] == a) return ym + m;
         else if(byy(a, ym[m])) r = m; else l = m + ;
     }
     if(l <  || l > N || ym[l].y != a.y) return NULL;
     return ym + l;
 }

 int main(){
     scanf("%d%*d%I64d", &N, &T); long long iT = T;
     f(i, , N){
         int cx, cy; char buf[];
         scanf("%d%d%s", &cx, &cy, buf);
         xm[i].x = ym[i].x = cx;
         xm[i].y = ym[i].y = cy;
         if(buf[] == '/'){
             xm[i].dr = ; ym[i].dr = ;
         }else{
             xm[i].dr = ; ym[i].dr = ;
         }
     }
     xm[] = (mir) {-, -, };
     ym[] = (mir) {-, -, };
     xm[N + ] = (mir) {, , };
     ym[N + ] = (mir) {, , };
     std::sort(xm + , xm + N + , byx);
     std::sort(ym + , ym + N + , byy);
     int cx = , cy = , cr = RR;
     for(;;){
         // printf("C %d %d = %d\n", cx, cy, cr);
         mir *res;
         if(cr == UU) res = findxmore((mir) {cx, cy + });
         else if(cr == LL) res = findyless((mir) {cx - , cy});
         else if(cr == DD) res = findxless((mir) {cx, cy - });
         else res = findymore((mir) {cx + , cy});
         if(res){
             if(cy ==  && res->y ==  && cx <  && res->x > ) T %= (iT - T - cx);
             int dis = abs(res->x - cx) + abs(res->y - cy);
             if(dis < T){
                 cx = res->x; cy = res->y; cr = (cr + res->dr) % ; T -= (long long) dis;
             }else{
                 printf("%I64d %I64d\n", (long long) cx + T * (long long) dx[cr], (long long) cy + T * (long long) dy[cr]);
                 return ;
             }
         }else{
             printf("%I64d %I64d\n", (long long) cx + T * (long long) dx[cr], (long long) cy + T * (long long) dy[cr]);
             return ;
         }
     }
     return ;
 }