2010年10月23日 总结

额，由于某种原因，导致昨天没写总结。今天的：

 

2^k进制数

采药

英语课句型转换

猪王争霸

提示问题

自然的谜语

扫雷

 

Fish学数学   之归并版

 

2^k进制数：比较难，涉及到组合数学及高精度。

program rq8;type arr=array[0..200] of longint;var k,w,maxn,i,tmp:longint; ans,b:arr;procedure plus(a,b:arr;var c:arr);var i:longint;begin fillchar(c,sizeof(c),0); if a[0]>b[0] then c[0]:=a[0] else c[0]:=b[0]; for i:=1 to c[0] do begin inc(c[i],a[i]+b[i]); inc(c[i+1],c[i] div 1000); c[i]:=c[i] mod 1000; end; if c[c[0]+1]<>0 then inc(c[0]);end;procedure mul(a:arr;b:longint;var c:arr);var i:longint;begin fillchar(c,sizeof(c),0); if b=0 then begin c[0]:=1;c[1]:=0;exit;end; c[0]:=a[0]; for i:=1 to c[0] do begin inc(c[i],a[i]*b); inc(c[i+1],c[i] div 1000); c[i]:=c[i] mod 1000; end; while c[c[0]+1]<>0 do begin inc(c[0]); inc(c[c[0]+1],c[c[0]] div 1000); c[c[0]]:=c[c[0]] mod 1000; end;end;procedure divi(a:arr;b:longint;var c:arr);var i,tmp:longint;begin fillchar(c,sizeof(c),0);c[0]:=a[0];tmp:=0; for i:= c[0] downto 1 do begin tmp:=tmp*1000+a[i]; c[i]:=tmp div b; tmp:=tmp mod b; end; while (c[0]>0)and(c[c[0]]=0) do dec(c[0]);end;procedure C(n,m:longint);var i:longint;beginfillchar(b,sizeof(b),0);b[0]:=1;b[1]:=1;for i:=n downto n-m+1 do mul(b,i,b);for i:=2 to m do divi(b,i,b);end;begin readln(k,w); maxn:=1; for i:=1 to k do maxn:=2*maxn; ans[0]:=0; for i:=2 to w div k do begin C(maxn-1,i); plus(ans,b,ans); end; tmp:=1; for i:=1 to w mod k do tmp:=tmp*2; dec(tmp); for i:=1 to tmp do//统计w div k+1位的2^k进制数的个数 begin C(maxn-1-i,w div k+1-1); plus(ans,b,ans); end; write(ans[ans[0]]); for i:=ans[0]-1 downto 1 do if ans[i]<10 then write('00',ans[i]) else if ans[i]<100 then write('0',ans[i]) else write(ans[i]); writeln;end.

采药：滥竽充数，无视之。

program rq15;var n,i,j,v:longint; c,w:array[1..100]of longint; f:array[0..1000]of longint;begin readln(v,n); for i:=1 to n do readln(c[i],w[i]); for i:=1 to n do for j:=v downto c[i] do if f[j-c[i]]+w[i]>f[j] then f[j]:=f[j-c[i]]+w[i]; writeln(f[v]);end.

英语课句型转换：把一个陈述句改写为感叹句，提取出系动词就可以了，注意主语是可以是数字的。

program rq146;var s,g,z,b,be:ansistring; bl,br,l,i:longint;begin readln(g); readln(s); l:=length(s)-1; for i:=1 to l do begin if (copy(s,i,4)=' is ')or(copy(s,i,4)=' am ') then begin be:=copy(s,i,3); bl:=i; br:=i+3; end; if copy(s,i,5)=' are ' then begin be:=copy(s,i,4); bl:=i; br:=i+4; end; end; z:=copy(s,1,bl-1); if (ord(z[1])>=65)and(ord(z[1])<=90) then z[1]:=chr(ord(z[1])+32); b:=copy(s,br+1,l-br); write(g,' ',b,' ',z,be,'!'); writeln;end.

猪王争霸：涉及高精度的排序和累加，还可以，就是写得ws了些。

program rq168;type pig=record m:string; da:ansistring; end; xx=array[0..10000] of longint;var n,i,kk,j,n1,n2:longint; bb:longint; a,b,c:xx; p:array[0..1001] of pig; t:pig; ans:longint; sum:ansistring;procedure sort(l,r:longint);var i,j:longint; m:ansistring; t:pig;begin i:=l; j:=r; m:=p[(l+r) div 2].da; repeat while p[i].da>m do inc(i); while m>p[j].da do dec(j); if i<=j then begin t:=p[i]; p[i]:=p[j]; p[j]:=t; inc(i); dec(j); end;until i>j; if l<j then sort(l,j); if i<r then sort(i,r);end;procedure cut(st:ansistring;t:longint;var d:xx);var tt:ansistring;l:longint;begin repeat l:=length(st); tt:=copy(st,l-7,8); val(tt,d[t]); dec(t); delete(st,l-7,8); until t=1; val(st,d[t]);end;procedure add;var ii,jj,yy,hh:longint;begin if bb>490 then begin for ii:=1 to n do delete(p[ii].da,1,bb-490); bb:=490; end; kk:=bb div 8+1; if kk>1 thenbegin cut(p[1].da,kk,a); for ii:=2 to n do begin cut(p[ii].da,kk,b); yy:=0; for jj:=kk downto 1 do begin c[jj]:=a[jj]+b[jj]+yy; yy:=c[jj] div 100000000; c[jj]:=c[jj] mod 100000000; end; for hh:=1 to kk do a[hh]:=c[hh]; if (c[kk+1]<>0)and((kk+1)*8<=490) then begin inc(kk);a[kk]:=c[kk];end; end;endelse for ii:=1 to n do begin val(p[ii].da,a[ii]); ans:=ans+a[ii]; end;end;begin readln(n); for i:=1 to n do begin readln(p[i].m);readln(p[i].da); end; for i:=2 to n do if length(p[i].da)>length(p[1].da) then begin t:=p[i]; p[i]:=p[1]; p[1]:=t; end; bb:=length(p[1].da); for i:=2 to n do while length(p[i].da)<length(p[1].da) do p[i].da:='0'+p[i].da; sort(1,n); repeat j:=i; while p[j].da=p[j-1].da do dec(j); for n1:=j to i-1 do for n2:=n1+1 to i do if p[n1].m>p[n2].m then begin t:=p[n1];p[n1]:=p[n2];p[n2]:=t; end; i:=j; dec(i); until i=0; for i:=1 to n do writeln(p[i].m); add; if ans<>0 then begin str(ans,sum); for i:=1 to 490-length(sum) do write('0'); writeln(sum); end else begin str(c[1],sum); if kk=62 then begin for i:=length(sum)-1 to length(sum) do write(sum[i]); end else begin for i:=1 to 490-length(sum)-(kk-1)*8 do write('0'); write(sum); end; for i:=2 to kk do begin if c[i]<10000000 then write('0'); if c[i]<1000000 then write('0'); if c[i]<100000 then write('0'); if c[i]<10000 then write('0'); if c[i]<1000 then write('0'); if c[i]<100 then write('0'); if c[i]<10 then write('0'); write(c[i]); end; end;end.

提示问题：比较简单但是写出来特别ws的字符串处理。

program rq212;var s,ans,f:string; i,j,l,k,sum:integer; b:boolean;begin readln(s); ans:=s; sum:=0; for i:=1 to length(s) do if ((ord(s[i])>=65)and(ord(s[i])<=90))or((ord(s[i])>=97)and(ord(s[i])<=122))then begin ans[i]:='.'; inc(sum); end; writeln(ans); l:=round(sum/3);j:=l; for i:=1 to length(s) do begin if ((ord(s[i])>=65)and(ord(s[i])<=90))or((ord(s[i])>=97)and(ord(s[i])<=122))then begin ans[i]:=s[i]; dec(j); end; if j=0 then break; end; k:=i+1; writeln(ans); b:=true; for i:=k to length(s) do if ((ord(s[i])>=65)and(ord(s[i])<=90))or((ord(s[i])>=97)and(ord(s[i])<=122)) then if (ord(s[i])=65)or(ord(s[i])=69)or(ord(s[i])=73)or(ord(s[i])=79)or(ord(s[i])=85)or(ord(s[i])=97)or(ord(s[i])=101)or(ord(s[i])=105)or(ord(s[i])=111)or(ord(s[i])=117) then begin b:=false; ans[i]:=s[i]; end; if b then begin l:=round(sum/3); j:=round((sum/3)*2)-l; for i:=k to length(s) do begin if ((ord(s[i])>=65)and(ord(s[i])<=90))or((ord(s[i])>=97)and(ord(s[i])<=122)) then begin ans[i]:=s[i]; dec(j); end; if j=0 then break; end; end; writeln(ans);end.

自然的谜语：标准的KMP，于是知道今天我才学会KMP...

program rq382;var i,j,k,l,n:longint; s1,s2:ansistring; a,next:array[1..100000] of longint;procedure main;var i,j,l1,l2:longint;begin i:=1;j:=1;n:=1; l1:=length(s1); l2:=length(s2); while i<=l2 do begin if (j=0) or (s2[i]=s1[j]) then begin inc(i); inc(j); end else j:=next[j]; if j>l1 then begin a[n]:=i-j+1; inc(n); j:=next[j]; end; end;end;begin readln(s1); readln(s2); l:=length(s1); j:=1; k:=0; while j<=l do begin if (k=0) or (s1[j]=s1[k]) then begin inc(j); inc(k); next[j]:=k; end else k:=next[k]; end; main; dec(n); if n=0 then write('There must be something wrong.') else begin writeln(n); for i:=1 to n do writeln(a[i]); end;end.

扫雷：额，简单模拟，输出即可。

program rq484;var m,n,i,j,x:longint; a:Array[0..101,0..101]of char; ans:array[1..100,1..100]of longint;function calc(i,j:longint):longint;begin calc:=0; if a[i-1,j]='*'then inc(calc); if a[i-1,j-1]='*'then inc(calc); if a[i-1,j+1]='*'then inc(calc); if a[i+1,j]='*'then inc(calc); if a[i+1,j-1]='*'then inc(calc); if a[i+1,j+1]='*'then inc(calc); if a[i,j-1]='*'then inc(calc); if a[i,j+1]='*'then inc(calc);end;begin readln(n,m); x:=0; while m*n<>0 do begin inc(x); for i:=1 to n do begin for j:=1 to m do read(a[i,j]); readln; end; for i:=1 to n do for j:=1 to m do if a[i,j]='.' then ans[i,j]:=calc(i,j); writeln('Field #',x,':'); for i:=1 to n do begin for j:=1 to m do if a[i,j]='*'then write('*')else write(ans[i,j]); writeln; end; readln(n,m); if m*n<>0 then writeln; end;end.

Fish学数学：标准的归并排序，排的时候就统计处逆序对数了。

program rq173;var i,n,ans:longint; a:array[1..20000]of longint;procedure merge(l,x,r:integer);var i,j,p:integer; b:array[1..20000]of longint;begin i:=l; j:=x+1; p:=l; while p<=r do begin if (i<=x)and((j>r)or(a[i]<=a[j])) then begin b[p]:=a[i]; inc(i); end else begin b[p]:=a[j]; inc(j); ans:=ans+x-i+1; end; inc(p); end; for i:=l to r do a[i]:=b[i];end;procedure msort(l,r:integer);var x:integer;begin if l<>r then begin x:=(l+r) div 2; msort(l,x); msort(x+1,r); merge(l,x,r); end;end;begin readln(n); for i:=1 to n do read(a[i]); ans:=0; msort(1,n); writeln(ans);end.

 

啊啊写完了，今天要早点睡了。

2010年10月23日23:02:16