USACO Section 1.1 题解

 为了备战2011NOIp，我决定把USACO重新刷过，练代码感觉，顺便回顾以前学的算法。

我觉得USACO的题十分好十分强大十分给力，每做一次感觉都不同，比赛前刷刷USACO是最好的选择。

我最支持USACO这种进阶模式，像我刷RQNOJ时，毫无目的的刷题，刷到哪算哪，不会的跳过，搞的我一点激情都没...

记得以前刷USACO时，十分兴奋，基本没都久，第一章就刷完了，貌似之前快刷到第4章了，貌似被卡住了，之后就很久没做了

虽然，现在刷USACO没以前那么积极了，刷玩一题就想放松下玩玩，不过这样也好，反正是为了准备比赛，慢慢来嘛...其实有些题做过再做还是有提高的（说不定你曾看过题解，现在自己想出来了，这不是很大的提高吗？我自己貌似之前也看过题解吧....+.+）

 

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这一节主要考查代码能力，不详细说了....

 

 

USACO 1.1.1     Your Ride Is Here 你要乘坐的飞碟在这里

 

program ride(input,output);const  a:array['A'..'Z'] of integer=(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26);var  m1,m2,i:longint;  s1,s2:string;procedure init; begin   assign(input,'ride.in');   assign(output,'ride.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;begin  init;  readln(s1);readln(s2);  m1:=1;m2:=1;  for i:=1 to length(s1) do   begin     m1:=m1*a[s1[i]];     if m1>47 then m1:=m1 mod 47;   end;  for i:=1 to length(s2) do   begin     m2:=m2*a[s2[i]];     if m2>47 then m2:=m2 mod 47;   end;  m1:=m1 mod 47; m2:=m2 mod 47;  if m1=m2 then writeln('GO')           else writeln('STAY');  outit;end.

 

 

USACO 1.1.2 Greedy Gift Givers 贪婪的礼物送礼者

 

program gift1(input,output);type   data=record       money:longint;       name:string;     end;var  n,i,j,x,y,k:longint;  s:string;  a:array[1..100] of data;procedure init; begin   assign(input,'gift1.in');   assign(output,'gift1.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;begin  init;  readln(n);  for i:=1 to n do readln(a[i].name);  for i:=1 to n do    begin      readln(s);      readln(x,y);      if (x<>0)and(y<>0) then        begin          k:=1;          while s<>a[k].name do inc(k);          dec(a[k].money,x-x mod y);        end;          for j:=1 to y do            begin              readln(s);              if x<>0 then               begin                k:=1;                while s<>a[k].name do inc(k);                inc(a[k].money,x div y);               end;            end;    end;  for i:=1 to n do writeln(a[i].name,' ',a[i].money);  outit;end.

 

 

USACO 1.1.3 Friday the Thirteenth 黑色星期五

 

program friday(input,output);const month:array[0..1,1..12] of integer=((3,0,3,2,3,2,3,3,2,3,2,3),(3,1,3,2,3,2,3,3,2,3,2,3));    //月份都减去28了，0是平年，1是闰年var a:array[0..6] of integer; i,k,j,h,n:longint;procedure init;begin assign(input,'friday.in'); assign(output,'friday.out'); reset(input); rewrite(output);end;procedure outit;begin close(input); close(output);end;begin init; readln(n);h:=6; for i:=1900 to 1900+n-1 do  begin   if (i mod 400=0) or ((i mod 100<>0) and (i mod 4=0)) then k:=1 else k:=0;   for j:=1 to 12 do    begin     inc(a[h]);     h:=(h+month[k,j]) mod 7;    end;  end; writeln(a[6],' ',a[0],' ',a[1],' ',a[2],' ',a[3],' ',a[4],' ',a[5]); outit;end.

 

 

USACO 1.1.4Broken Necklace破碎的项链

 

program beads(input,output);var  n,i,max,ans,k:longint;  ch:char;  s:ansistring;               //在pascal中string只有255个字符，要用ansistring，其他语言不知道到要不要注意procedure init;begin assign(input,'beads.in'); assign(output,'beads.out'); reset(input); rewrite(output);end;procedure outit;begin close(input); close(output);end;begin  init;  readln(n);  readln(s);  s:=s+s+s;  for i:=n+1 to n+n do    begin      ans:=0;      k:=i-1;                                       //left      ch:='w';      while (ch='w')or(ch=s[k])or(s[k]='w') do        begin          inc(ans);          if s[k]<>'w' then ch:=s[k];          dec(k);        end;      k:=i;                                         //right      ch:='w';      while (ch='w')or(ch=s[k])or(s[k]='w') do        begin          inc(ans);          if s[k]<>'w' then ch:=s[k];          inc(k);        end;      if ans>max then max:=ans;      if ans>n then begin max:=n; break;end;    end;  writeln(max);  outit;end.