2007 矩阵取数把不用高精度的程序发给我好么.今天6点以前.我会尽我所能多给分的pascal语言.c++我看不懂.NO

2007 矩阵取数
把不用高精度的程序发给我好么.
今天6点以前.
我会尽我所能多给分的
pascal语言.
c++我看不懂.
NOIP2007提高组复赛第三题.

不用高精度是不行的,不能够AC,因为就算QWORD也装不下.
【pascal程序给你了.带题解】
【题解】
显然取数的规则决定了他有最优子结构,即取了一个数后,剩下的又可以看成是规模更小的子问题
而且每一排的问题性质是一样的,我就单看一排
设f[I,j]是按规则计分的情况下从i到j这一段能取得的最大值,
则：f[I,j]=max
{a[I]*2+f[i+1,j]*2,
F[I,j-1]*2+a[j]*2}
初始值f[I,i]=a[i]*2
优化的时候可以把方程里面的2提出来,毕竟是高精度运算,能少算一点就少算一点,
【程序】
program game;
const mn=80;
nn=10000;
type nt=array [0..100] of longint;
ntt=^nt;
var a:array [1..mn,1..mn] of longint;
f:array [1..mn,1..mn,1..mn] of {longint}ntt;
n,m:integer;
procedure rd;
var i,j:integer;
begin
readln(n,m);
fillchar(a,sizeof(a),0);
for i:=1 to n do
begin
for j:=1 to m do read(a[i,j]);
readln;
end;
end;
procedure two(var a,b:nt);
var i,l:integer;
x:longint;
begin
l:=b[0];
fillchar(a,sizeof(a),0);
x:=0;
for i:=1 to l do
begin
x:=x+b[i]*2;
a[i]:=x mod nn;
x:=x div nn;
end;
if x>0 then begin inc(l); a[l]:=x end;
a[0]:=l;
end;
procedure plus1(var a,b:nt;t:longint);
var i:integer;
begin
a:=b;
inc(a[1],t);
i:=1;
for i:=1 to a[0] do
begin
inc(a[i+1],a[i] div nn);
a[i]:=a[i] mod nn;
end;
if a[i+1]0 then inc(a[0]);
end;
procedure plus2(var a,c:nt);
var l,i:integer;
begin
if a[0]>c[0] then l:=a[0] else l:=c[0];
for i:=1 to l do
begin
inc(a[i],c[i]);
inc(a[i+1],a[i] div nn);
a[i]:=a[i] mod nn;
end;
if a[l+1]0 then inc(l);
a[0]:=l;
end;
function ma(a,b:nt):integer;
var i:longint;
begin
if a[0]>b[0] then exit(1);
if a[0]1) and (a[i]=b[i]) do dec(i);
if a[i]>b[i] then exit(1) else exit(2);
end;
end;
procedure dp;
var d,tt,i,j,k:integer;
t:longint;
sum,t1,t2:nt;
procedure fill(var a:nt);
begin fillchar(a,sizeof(a),0); end;
begin
for i:=1 to n do
for j:=1 to m do
begin
new(f[i,j,j]);
fill(f[i,j,j]^);
f[i,j,j]^[0]:=1;
f[i,j,j]^[1]:=a[i,j]*2;
end;
for d:=1 to m-1 do
for i:=1 to m-d do
begin
j:=i+d;
for k:=1 to n do
begin
plus1(t1,f[k,i+1,j]^,a[k,i]);
plus1(t2,f[k,i,j-1]^,a[k,j]);
tt:=ma(t1,t2);
new(f[k,i,j]);
if tt=1 then two(f[k,i,j]^,t1)
else if tt=2 then two(f[k,i,j]^,t2);
end;
end;
fillchar(sum,sizeof(sum),0);
sum[0]:=1;
for i:=1 to n do
plus2(sum,f[i,1,m]^);
write(sum[sum[0]]);
for t:=sum[0]-1 downto 1 do
begin
write(sum[t] div 1000);
write(sum[t] div 100 mod 10);
write(sum[t] div 10 mod 10);
write(sum[t] mod 10);
end;
end;
begin
assign(input,'game.in');reset(input);
assign(output,'game.out');rewrite(output);
rd;
dp;
close(input); close(output);
end.