讨论 / ge wei da niu kuai lai bang wo a
许子望 2012-01-10 21:02:00
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yin wei xu ni qi wu fa shi yong han yu,wo zhe hao yong pin yin

zai xia bei (jin min de yu suan fang an)zhe dao ti ge lei dao le pian pian shuo wo zhi de le 20 fen

xi wang ge wei da niu duo duo bang zhu

xia mian shi yuan dai ma

uses math;

var a:array[1..60,1..3]of longint;

b:array[1..60,0..2]of longint;

f:array[0..3200]of longint;

i,j,n,v,p:longint;

begin

readln(v,n);

v:=v div 10;

fillchar(b,sizeof(b),0);

for i:=1 to n do begin

read(a[i,1],a[i,2],a[i,3]);

readln;

a[i,1]:=a[i,1] div 10;

if a[i,3]<>0 then begin

inc(b[a[i,3],0]);

b[a[i,3],b[a[i,3],0]]:=i;

end;

end;

for i:=1 to n do

if a[i,3]=0 then begin

for j:=v downto a[i,1] do

f[j]:=max(f[j],f[j-a[i,1]]+a[i,2]*a[i,1]);

if b[i,1]<>0 then

for j:=v downto a[i,1]+a[b[i,1],1] do

f[j]:=max(f[j],f[j-a[i,1]-a[b[i,1],1]+a[i,2]*a[i,1]+a[b[i,1],1]*a[b[i,1],2]]);

if b[i,2]<>0 then begin

for j:=v downto a[i,1]+a[b[i,2],1] do

f[j]:=max(f[j],f[j-a[i,1]-a[b[i,2],1]]+a[i,1]*a[i,2]+a[b[i,2],1]*a[b[i,2],2]);

for j:=v downto a[i,1]+a[b[i,2],1]+a[b[i,1],1] do

f[j]:=max(f[j],f[j-a[b[i,1],1]-a[b[i,2],1]-a[i,1]]+a[i,1]*a[i,2]+a[b[i,1],1]*a[b[i,1],2]+a[b[i,2],1]*a[b[i,2],2]);

end;

end;

write(10*f[v]);

end.

bai tuo ge wei le

#1 lijiaming12340@2011-12-03 19:39:00
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你会写汉字?
#2 jackchen1999@2012-01-10 21:02:00
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楼主拼音不太好。。。

如题

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