NOIP2012
DAY1
Vigenère 密码
模拟
可以先把密钥全部转化为大写或小写
1 var 2 i,asc:longint; 3 k,c:ansistring; 4 begin 5 readln(k); 6 for i:=1 to length(k) do k[i]:=upcase(k[i]); 7 readln(c); 8 while length(k)<length(c) do k:=k+k; 9 for i:=1 to length(c) do 10 begin 11 asc:=ord(c[i])-(ord(k[i])-ord('A')); 12 if (ord(c[i])>=65) and (ord(c[i])<=90) then 13 if asc<65 then asc:=ord('Z')-(65-asc)+1; 14 if (ord(c[i])>=97) and (ord(c[i])<=122) then 15 if asc<97 then asc:=ord('z')-(97-asc)+1; 16 write(chr(asc)); 17 end; 18 end.
1 var 2 k,m,c:ansistring; 3 i,l,s:longint; 4 dax1,xix1,dax2,xix2:boolean; 5 procedure pd(x,y:char); 6 begin 7 dax1:=false; dax2:=false; 8 xix1:=false; xix2:=false; 9 if (ord(x)>=ord('a'))and(ord(x)<=ord('z')) then xix1:=true 10 else if (ord(x)>=ord('A'))and(ord(x)<=ord('Z')) then dax1:=true; 11 if (ord(y)>=ord('a'))and(ord(y)<=ord('z')) then xix2:=true 12 else if (ord(y)>=ord('A'))and(ord(y)<=ord('Z')) then dax2:=true; 13 end; 14 procedure doit1(var k,c:char); 15 var cha:longint; 16 begin 17 cha:=ord(c)-(ord(k)-ord('a')); 18 if cha<ord('a') then 19 cha:=ord('z')-((ord('a')-cha))+1; 20 c:=chr(cha); 21 end; 22 procedure doit2(var k,c:char); 23 var cha:longint; 24 begin 25 cha:=ord(c)-(ord(k)-ord('A')); 26 if cha<ord('A') then 27 cha:=ord('Z')-((ord('A')-cha))+1; 28 c:=chr(cha); 29 end; 30 procedure doit3(var k,c:char); 31 var cha:longint; 32 begin 33 cha:=ord(c)-(ord(k)-ord('a')); 34 if cha<ord('A') then 35 cha:=ord('Z')-((ord('A')-cha))+1; 36 c:=chr(cha); 37 end; 38 procedure doit4(var k,c:char); 39 var cha:longint; 40 begin 41 cha:=ord(c)-(ord(k)-ord('A')); 42 if cha<ord('a') then 43 cha:=ord('z')-((ord('a')-cha))+1; 44 c:=chr(cha); 45 end; 46 begin 47 assign(input,'vigenere.in'); 48 reset(input); 49 assign(output,'vigenere.out'); 50 rewrite(output); 51 readln(k); 52 readln(c); 53 while l<length(c) do 54 begin 55 k:=k+k; 56 l:=length(k); 57 end; 58 for i:=1 to length(c) do 59 begin 60 pd(k[i],c[i]); 61 if (xix1)and(xix2) then doit1(k[i],c[i]); 62 if (dax1)and(dax2) then doit2(k[i],c[i]); 63 if (xix1)and(dax2) then doit3(k[i],c[i]); 64 if (dax1)and(xix2) then doit4(k[i],c[i]); 65 end; 66 writeln(c); 67 close(input); 68 close(output); 69 end.
1 注意大小写 2 var 3 i,l,t:longint; 4 a:Array[0..1000] of longint; 5 s:string; 6 ch:char; 7 begin 8 readln(s); 9 l:=length(s); 10 for i:=1 to l do 11 if ord(s[i])>=97 then a[i]:=ord(s[i])-96 12 else a[i]:=ord(s[i])-64; 13 a[0]:=a[l]; 14 read(ch); 15 i:=1; 16 t:=ord(ch); 17 while (t>64)and(t<124) do 18 begin 19 if t<97 then 20 begin 21 t:=t-64; 22 t:=(t-a[i mod l]+27) mod 26; 23 if t=0 then t:=26; 24 t:=t+64; 25 write(chr(t)); 26 end 27 else 28 begin 29 t:=t-96; 30 t:=(t-a[i mod l]+27) mod 26; 31 if t=0 then t:=26; 32 t:=t+96; 33 write(chr(t)); 34 end; 35 read(ch); 36 inc(i); 37 t:=ord(ch); 38 end; 39 end.
智商在流失。。
国王游戏
快排+高精乘+高精除
[转]
正解= 贪心+高精.
贪心证明 :
设存在 相邻大臣A(l1,r1),B(l2,r2),A,B前左手乘积为Sum :
当A在B前时:
则Ans=max(Sum/r1,Sum*l1/r2) ;
当B在A前时:
则Ans=max(Sum/r2,Sum*l2/r1) ;
显然 Sum*l2/r1>Sum/r1 ;
Sum*l1/r2>Sum/r2 ;
所以当 Sum*l2/r1>Sum*l1/r2
即 l2*r2>l1*r1 时 A应在B前
同理
即 l2*r2<l1*r1 时 B应在A前
所以按 l*r 从小到大排序在进行模拟即为正解.
1 type 2 arr=array[1..10000] of longint; 3 var 4 ans:arr; 5 n,i,j,a1,b1,la:longint; 6 a,b,c:array[1..10000] of longint; 7 procedure sort(l,r:longint); 8 var 9 i,j,tmp,mid:longint; 10 begin 11 i:=l; 12 j:=r; 13 mid:=c[(i+j)div 2]; 14 repeat 15 while c[i]<mid do inc(i); 16 while c[j]>mid do dec(j); 17 if i<=j then 18 begin 19 tmp:=c[i]; c[i]:=c[j]; c[j]:=tmp; 20 tmp:=a[i]; a[i]:=a[j]; a[j]:=tmp; 21 tmp:=b[i]; b[i]:=b[j]; b[j]:=tmp; 22 inc(i); 23 dec(j); 24 end; 25 until i>j; 26 if l<j then sort(l,j); 27 if i<r then sort(i,r); 28 end; 29 procedure mul(var a:arr;c:longint;var la:longint); 30 var i:longint; 31 begin 32 a[1]:=a[1]*c; 33 for i:=2 to la do 34 begin 35 a[i]:=a[i]*c; 36 a[i]:=a[i]+a[i-1] div 10; 37 a[i-1]:=a[i-1] mod 10; 38 end; 39 while a[la]>=10 do 40 begin 41 inc(la); 42 a[la]:=a[la-1] div 10; 43 a[la-1]:=a[la-1] mod 10; 44 end; 45 end; 46 procedure chufa(var d:arr; b:longint; var la:longint); 47 var c,a:arr; 48 x,j,tmp:longint; 49 begin 50 fillchar(c,sizeof(c),0); 51 for i:=la downto 1 do a[la-i+1]:=d[i]; 52 //for i:=1 to la do write(a[i]); writeln; 53 x:=0; 54 for i:=1 to la do begin 55 c[i]:=(x*10+a[i]) div b; 56 x:=(x*10+a[i]) mod b; end; 57 j:=1; 58 while (c[j]=0) and (j<la) do inc(j); 59 if (j=la) and (c[la]=0) then begin writeln(1); halt; end; 60 for i:=j to la do 61 write(c[i]); 62 end; 63 begin 64 assign(input,'kinggame.in'); 65 reset(input); 66 assign(output,'kinggame.out'); 67 rewrite(output); 68 readln(n); 69 readln(a1,b1); 70 for i:=1 to n do 71 begin 72 readln(a[i],b[i]); 73 c[i]:=a[i]*b[i]; 74 end; 75 sort(1,n); 76 fillchar(ans,sizeof(ans),0); 77 ans[1]:=1; la:=1; 78 for i:=1 to n-1 do mul(ans,a[i],la); 79 mul(ans,a1,la); 80 //for i:=la downto 1 do write(ans[i]); writeln(b[n]); 81 chufa(ans,b[n],la); 82 //for i:=la downto 1 do write(ans[i]); 83 //for i:=1 to n do writeln(a[i],' ',b[i],' ',c[i]); 84 close(input); 85 close(output); 86 end.
1 可以证明最优序列是按照左手上的数*右手上的数递增排序，所以 2 1. 按照左手上的数*右手上的数递增排序 3 2. 前n-1个左手上的数累乘（高精乘） 4 3. 除以n右手上的数（高精除） 5 var 6 i,n,l:longint; 7 a,b,c:array[0..10000] of longint; 8 g,g2:array[0..100000] of longint; 9 procedure gj1; 10 var 11 j:longint; 12 begin 13 for j:=1 to l do g[j]:=g[j]*b[i]; 14 for j:=1 to l do 15 begin 16 g[j+1]:=g[j+1]+g[j] div 10; 17 g[j]:=g[j] mod 10; 18 end; 19 inc(l); 20 while g[l]>9 do 21 begin 22 g[l+1]:=g[l+1]+g[l] div 10; 23 g[l]:=g[l] mod 10; 24 inc(l); 25 end; 26 if g[l]=0 then dec(l); 27 end; 28 procedure gj2; 29 var 30 j:longint; 31 begin 32 for j:=l downto 1 do 33 begin 34 g[j-1]:=g[j-1]+(g[j] mod c[n])*10; 35 g[j]:=g[j] div c[n]; 36 end; 37 while g[l]=0 do dec(l); 38 end; 39 procedure sort(l,r:longint); 40 var 41 i,j,x,y:longint; 42 begin 43 i:=l; 44 j:=r; 45 x:=a[(l+r) div 2]; 46 repeat 47 while a[i]<x do inc(i); 48 while x<a[j] do dec(j); 49 if not (i>j) then 50 begin 51 y:=a[i]; 52 a[i]:=a[j]; 53 a[j]:=y; 54 y:=b[i]; 55 b[i]:=b[j]; 56 b[j]:=y; 57 y:=c[i]; 58 c[i]:=c[j]; 59 c[j]:=y; 60 inc(i); 61 dec(j); 62 end; 63 until (i>j); 64 if l<j then sort(l,j); 65 if i<r then sort(i,r); 66 end; 67 begin 68 readln(n); 69 readln(b[0],c[0]); 70 for i:=1 to n do 71 begin 72 read(b[i],c[i]); 73 a[i]:=b[i]*c[i]; 74 end; 75 sort(1,n); 76 l:=1; 77 g[1]:=b[0]; 78 for i:=1 to n-1 do gj1; 79 gj2; 80 for i:=l downto 1 do write(g[i]); 81 writeln; 82 end.
