ac1212.pas
来自「某牛人写的acm.tongji.edu.cn上大部分ac的代码,仅供学习研究,请」· PAS 代码 · 共 40 行
PAS
40 行
program tju1212;
const
maxd=30;
maxl=10;
modulo=11380;
var
f:array[0..maxd,0..maxl,0..maxl,0..maxl]of word;
d,l1,l2,l3,i,j,k,t:longint;
begin
f[0,0,0,0]:=1;
for d:=1 to maxd do
for l1:=0 to maxl do
for l2:=0 to maxl do
for l3:=0 to maxl do begin
if l1+l2+l3=0 then begin f[d,l1,l2,l3]:=1;continue;end;
t:=0;
if l3>0 then
for i:=0 to l3-1 do
t:=(t+f[d-1,0,0,i]*f[d,l1,l2,l3-1-i]) mod modulo;
if l2>0 then
for i:=0 to l2-1 do
for j:=0 to l3 do
t:=(t+f[d-1,0,i,j]*f[d,l1,l2-1-i,l3-j]) mod modulo;
if l1>0 then
for i:=0 to l1-1 do
for j:=0 to l2 do
for k:=0 to l3 do
t:=(t+f[d-1,i,j,k]*f[d,l1-1-i,l2-j,l3-k]) mod modulo;
f[d,l1,l2,l3]:=t;
end;
repeat
read(i,j,k,d);
if d>0 then
writeln((f[d,i,j,k]-f[d-1,i,j,k]+modulo) mod modulo)
else
writeln(ord(i+j+k=0));
until seekeof;
end.
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