各位大牛。。。

这道题我是用最朴素的算法...

f表示左边开始取了i次.右边开始取了j次的最大值....

f:=max(f+a[i]*2^(i+j),f+a[j]*2^(i+j))

为什么我会错了啊?!?!????

const

maxm=81;

maxn=50;

mk=1000000;

type

sty=array[0..maxn] of longint;

var

ans:sty;

pdf:int64;

n,m:longint;

i,j,k,t:longint;

twice:array[0..maxm] of sty;

data:array[0..maxm] of longint;

f:array[-1..maxm,-1..maxm] of sty;

procedure print;

var

s:string;

i,j,k:longint;

begin

if ans[0]=1 then writeln(ans[1]) else begin

write(ans[ans[0]]);

for i:=ans[0]-1 downto 1 do begin

str(ans[i],s);

for j:=1 to (6-length(s)) do write('0');

write(ans[i]);

end;

end;

close(input);

close(output);

end;

function mak(x,y:longint):longint;

begin

if x>y then exit(x)

else exit(y);

end;

function max(x,y:sty):sty;

var

flag:boolean;

i,j,k:longint;

begin

if x[0]>y[0] then exit(x);

if y[0]>x[0] then exit(y);

flag:=true;

for i:=x[0] downto 1 do if y[i]>x[i] then begin

flag:=false;

break;

end;

if flag then exit(x)

else exit(y);

end;

procedure inq(x:sty;var y:sty);

var

c:sty;

len:longint;

i,j,k:longint;

begin

k:=0;

len:=mak(x[0],y[0]);

fillchar(c,sizeof(c),0);

for i:=1 to len do begin

c[i]:=(x[i]+y[i]+k) mod mk;

k:=(x[i]+y[i]+k) div mk;

end;

c[0]:=len;

if k>0 then begin

inc(c[0]);

c[c[0]]:=k;

end;

y:=c;

end;

procedure prem(x,y:longint;var z:sty);

var

i,j,k:longint;

begin

fillchar(z,sizeof(z),0);

for i:=1 to twice[y,0] do begin

pdf:=twice[y,i]*x+z[i];

z[i]:=pdf mod mk;

z:=pdf div mk;

end;

z[0]:=twice[y,0];

if z[z[0]+1]0 then inc(z[0]);

while (z[z[0]]=0)and(z[0]0) do dec(z[0]);

end;

procedure refil;

var

i,j,k:longint;

begin

twice[0,0]:=1;

twice[0,1]:=1;

for i:=1 to maxm do begin

k:=0;

twice:=twice;

for j:=1 to twice do begin

twice:=(twice*2+k) mod mk;

k:=(twice*2+k) div mk;

end;

if k0 then begin

inc(twice);

twice[i,twice]:=k;

end;

end;

end;

procedure work;

var

a,b,tot:sty;

begin

refil;

for t:=1 to n do begin

fillchar(tot,sizeof(tot),0);

for i:=1 to m do read(data[i]);

for i:=0 to m do begin

for j:=0 to m do begin

if i+j>m then break;

prem(data[i],i+j,a);

prem(data[m-j+1],i+j,b);

inq(f,a);

inq(f,b);

f:=max(a,b);

end;

end;

for i:=0 to m do tot:=max(tot,f);

inq(tot,ans);

readln;

end;

end;

procedure init;

begin

assign(input,'game.in');

reset(input);

assign(output,'game.out');

rewrite(output);

readln(n,m);

end;

begin

init;

work;

print;

end.