Jiezhi's Blog

好记性不如烂笔头

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情恨天
长剑耿介
长相思，在佳人。一日不见，思之如狂。狂兮狂兮可奈何？只恨相见.
既相见，何相识？芳草凋零，北雁南飞。飞兮飞兮留不住，自是流年。

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作业调度方案

本来想用记录p=record u,b,e:integer; end; （u为第几道工序，b为开始时间，e为结束时间）

来实现的，直接排还是挺方便的，但有些地方不好实现就没做下去……

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能量项链

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rogram energy;

{20:02 }

var

    a1,a2:array[1..200]of integer;

    q:array[1..200,1..200]of longint;

         i,j,n,k,m,p:integer;

         best:longint;

begin

assign(input,'energy.in');reset(input);

assign(output,'energy.ans');rewrite(output);

readln(n);

for i:=1 to n do read(a1[i]);

readln;

for i:=1 to n-1 do a2[i]:=a1[i+1];

a2[n]:=a1[1];

m:=n*2-1;

for i:=n+1 to m do begin

         a1[i]:=a1[i-n];a2[i]:=a2[i-n]

end;

//for i:=1 to m do writeln(a1[i],' ',a2[i],' ');

fillchar(q,sizeof(q),0);

for p:=1 to n-1 do

         for i:=1 to m-1 do

                   begin

                 j:=i+p;

                 if j>m then break;

                            for k:=i to j-1 do begin

                            if q[i,j]<q[i,k]+q[k+1,j]+a1[i]*a2[k]*a2[j]

                               then q[i,j]:=q[i,k]+q[k+1,j]+a1[i]*a2[k]*a2[j];

                            end;

               end;

best:=0;

for i:=1 to n do

         if best<q[i,i+n-1]then best:=q[i,i+n-1];

writeln(best);

close(input);close(output);

end.

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金明的预算方案

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program budget;
 
{20:03 }
 
type node=record
 
            u:integer;
 
       v,p:array[0..2]of integer;
 
     end;
 
var w:array[1..60]of node;
 
    f:array[0..60,0..2000]of longint;
 
    vx,px,qx,g:array[1..60]of integer;
 
    m,n,i,k,j:integer;
 
begin
 
assign(input,'budget.in');reset(input);
 
assign(output,'budget.out');rewrite(output);
 
readln(n,m);
 
k:=0;
 
for i:=1 to m do begin
 
      readln(px[i],vx[i],qx[i]);
 
      if qx[i]=0 then begin
 
                   inc(k);g[i]:=k;
 
                   with w[k] do begin
 
                   u:=0;
 
                   v[0]:=vx[i];p[0]:=px[i];
 
                   for j:=1 to 2 do begin
 
                            v[j]:=0;p[j]:=0;
 
                   end;
 
                   end;
 
         end;
 
    end;
 
for i:=1 to m do begin
 
         if qx[i]<>0 then
 
         with w[g[qx[i]]] do begin
 
                   inc(u);v[u]:=vx[i];p[u]:=px[i];
 
         end;
 
end;
 
for i:=1 to k do
 
         with w[i] do begin
 
                   write('w[',i,']:');
 
         for j:=0 to 2 do write('(',v[j],',',p[j],') ');
 
                   writeln;
 
end;
 
         
 
fillchar(f,sizeof(f),0);
 
         
 
for i:=1 to k do
 
         for j:=1 to n do
 
                   with w[i] do begin
 
             f[i,j]:=f[i-1,j];
 
                   
 
             if (f[i,j]< (f[i-1,j-p[0]]+p[0]*v[0]))and(j>=p[0]) then
 
                            f[i,j]:=f[i-1,j-p[0]]+p[0]*v[0];
 
                   
 
                   if (f[i,j]< (f[i-1,j-p[0]-p[1]]+p[0]*v[0]+v[1]*p[1]))and(j>=p[0]+p[1]) then
 
                            f[i,j]:=f[i-1,j-p[0]-p[1]]+p[0]*v[0]+v[1]*p[1];
 
                   
 
                   if (f[i,j]< (f[i-1,j-p[0]-p[2]]+p[0]*v[0]+p[2]*v[2]))and(j>=p[0]+p[2]) then
 
                            f[i,j]:=f[i-1,j-p[0]-p[2]]+p[0]*v[0]+p[2]*v[2];
 
                   
 
                   if (f[i,j]< (f[i-1,j-p[0]-p[1]-p[2]]+p[0]*v[0]+v[1]*p[1]+p[2]*v[2]))and(j>=p[0]+p[1]+p[2]) then
 
                            f[i,j]:=f[i-1,j-p[0]-p[1]-p[2]]+p[0]*v[0]+v[1]*p[1]+p[2]*v[2];
 
                   writeln('f[',i,',',j,']:',f[i,j]);
 
         end;
 
         writeln(f[k,n]);
 
         close(input);close(output);
 
end.

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就像解题报告中说的超简单，但还是出了许多小毛病,如刚开始将sum同avi,paper等定义在一起，而我为了省空间将它们限制在0..100，故算出来的sum错误无疑了；用字符读入数据时忘了空格……最后在比较时本想输出sum同最先出现的人却费事的用downto，结果没用‘<=’反画蛇添足了……唉……太粗心了……

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可以说超简单的一道题，学过数组的人都应会做

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program save;
var a:array[0..12]of integer;
    i,x,p:integer;
begin
assign(input,'save.in');reset(input);
assign(output,'');rewrite(output);
for i:=1 to 12 do a[i]:=300;
a[0]:=0;p:=0;
for i:=1 to 12 do begin
readln(x);inc(a[i],a[i-1]);dec(a[i],x);
     if a[i]<0 then begin
   writeln('-',i);
      close(input);close(output);exit;
end else {if a[i]>=100 then} begin
   p:=(a[i]div 100)*100;
   inc(a[0],p);
      a[i]:=a[i]-p;
end;
end;
a[0]:=a[0]*6 div 5;
writeln(a[0]+a[12]);
close(input);close(output);
end.

Posted on Edited on In Changyan: Disqus:

首先快排是要的，下面就是计算最小的体力耗费值了，其中我用到了冒泡排序，具体见下：

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program fruit;

var a:array [0..10001]of longint;

       i,j,k,n:integer;

          t,s:longint;

          flag:boolean;

procedure fastsort(l,r:integer);

var lx,rx,pi:integer;

begin

lx:=l;rx:=r;pi:=a[l];

while lx<rx do begin

    while a[rx]>pi do dec(rx);

    if lx<rx then begin

           t:=a[rx];a[rx]:=a[lx];a[lx]:=t;inc(lx);

    end;

    while a[lx]<pi do inc(lx);

    if lx<rx then begin

           t:=a[rx];a[rx]:=a[lx];a[lx]:=t;dec(rx);

    end;

end;

       if l<lx-1 then fastsort(l,lx-1);

       if r>rx+1 then fastsort(rx+1,r);

end;

begin

assign(input,'fruit.in');reset(input);

assign(output,'fruit.ans');rewrite(output);

readln(n);

for i:=1 to n do read(a[i]);

fastsort(1,n);

s:=0;a[0]:=0;

for i:=1 to n do begin

       s:=s+a[i]+a[i-1];a[i]:=a[i-1]+a[i];

              for k:=i to n-1 do

                     if a[k]>a[k+1] then begin

                            t:=a[k];a[k]:=a[k+1];a[k+1]:=t;

                     end;

end;

       dec(s,a[1]);

       writeln(s);

close(input);close(output);

end.

其中还有许多可以优化的地方，如：可以用t:=a[i];a[i]:=a[n];a[n]:=t;来减少后面的排序次数，又数组已是有序的了故还可在冒泡排序上优化，可置一布尔型变量flag来提前结束排序.

但十个测试点都过了就不做了.