秒了

program cheng;

var

f:array[1..40,1..6] of longint;

i,j,ans,k,m,n:longint;

a:array[1..40] of integer;

ss:string;

function yinzi(x,y:integer):longint;

var

i:integer;

begin

ans:=0;

for i:=x to y do

ans:=ans*10+a[i];

exit(ans);

end;

begin

readln(n,m);

readln(ss);

for i:=1 to n do

a[i]:=ord(ss[i])-48;

for i:=1 to n do

f:=yinzi(1,i);

for i:=2 to m+1 do

for j:=i to n do

for k:=j-1 downto 1 do

if f[k,i-1]*yinzi(k+1,j)>f[j,i] then

f[j,i]:=f[k,i-1]*yinzi(k+1,j);

writeln(f[n,m+1]);

end.

读入数组后，从最后一个数开始搜索吧，这样优化程序，减少时间与空间

写一堆for就够了

何必那么麻烦呢

han!!!!!!

#include

#include

#include

int n,k;

long a[50][10]={0},c[100]={0},f[100]={0},p[100]={0};

char d[50];

void init()

{int i;

scanf("%d%d\n",&n,&k);

for(i=1;i

var m,b,x,n,p,t,s,w,i:longint;

a:array[1..10000] of longint;

nn,aa,kk,xx:string;

begin

readln(n,p);

t:=1;

read(nn);

repeat

begin

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

if s>=n then

begin

if (s=n) and (t=p+1) then

begin

w:=1;aa:=copy(nn,w,a[1]);val(aa,b);

for i:=2 to t do

begin

w:=w+a;val(copy(nn,w,a[i]),x);b:=b*x;

val(copy(nn,w,a[i]),x);str(x,xx);aa:=aa+' *'+xx;

end;

if b>m then begin m:=b;kk:=aa;end;

end;

s:=s-a[t];t:=t-1;s:=s-a[t];

end

else begin t:=t+1;a[t]:=0;end;

end;

until t=0;

write(m);

end.

var

num:array[1..40,0..30]of longint;

a:array[1..40,0..30]of longint;

i,tt,j,n,m,x,man,code:longint;

s,ttt:string;

procedure change( i,j:longint; var p:string);

var

k:longint;

begin

p:='';

for k:=i to j do

p:=p+s[k];

end;

function max(l,r:longint):longint;

begin

if l>r then max:=l

else max:=r;

end;

begin

readln(n,m);

read(s);

fillchar(a,sizeof(a),0);

for i:=1 to n do

for j:=1 to n do

begin

tt:=0;

change(i,j,ttt);

val(ttt,tt,code);

num:=tt;

end;

for i:=1 to n do

a:=num[1,i];

for i:=1 to n do

for j:=1 to m do

for x:=1 to i-1 do

begin

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

end;

write(a[n,m]);

end.

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

var a,b:array[0..100,0..100] of longint;

i,j,k,n,m:longint;

s:string;

begin

readln(n,m);

readln(s);

for i:=1 to n do a:=ord(s[i])-ord('0');

for i:=1 to n do

for j:=i+1 to n do

begin

a:=a*10+(ord(s[j])-ord('0'));

a[j,i]:=a;

end;

for i:=1 to n do b:=a[1,i];

{for i:=1 to 50 do for j:=0 to 50 do if b=0 then b:=1;}

for j:=1 to m do

for i:=2 to n do

for k:=1 to i-1 do

if (b

var

st:string;

n,s,i,j,k:longint;

rec:array[0..50,0..50] of int64;

function cut(a,b:longint):longint;

var t:longint;

st1:string;

begin

st1:=copy(st,a,b-a+1);

val(st1,t);

exit(t);

end;

function search(a,b:longint):longint;

var

i,k,max,c:longint;

begin

max:=0;

if rec[a,b]-1 then begin exit(rec[a,b]);end;

for i:=a+1 to s-b+1 do

begin

k:=cut(a,i-1)*search(i,b-1);

if k>max then begin max:=k; c:=i;end;

end;

rec[a,b]:=max;

exit(max);

end;

begin

fillchar(rec,sizeof(rec),-1);

readln(s,n);

readln(st);

for i:=1 to s do

rec:=cut(i,s);

writeln(search(1,n));

end.

数据弱,深搜也过得了.longint就够,才4个测试点

• -

大牛 看看 谢谢

var

xx:qword;

s,s1:string;

n,k,i,j,l:longint;

f:array[1..40,0..30] of qword;

begin

readln(n,k);

readln(s);

f[1,1]:=ord(s[1])-48;

for l:=1 to k do

for i:=1 to n do

for j:=1 to i-1 do

if l0

then xx:=f[j,l-1]*xx;

if f

program p1347;

var

o:array[1..40,1..7] of qword;

num:array[1..40,1..40] of qword;

n,k,code,i,j,x:integer;

s,s1:string;

function max(a,b:qword):qword;

begin

if a>b then exit(a) else exit(b);

end;

begin

readln(n,k);

readln(s);

fillchar(num,sizeof(num),0);

for i:=1 to n do

for j:=i to n do

begin

s1:=copy(s,i,j-i+1);

val(s1,num,code);

end;

for i:=1 to n do

o:=num;

for i:=n downto 1 do

for j:=2 to k+1 do

begin

for x:=n-j+1 downto i do

o:=max(o,num*o[x+1,j-1]);

end;

writeln(o[1,k+1]);

end.

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

巨水！巨水！巨水！巨水！巨水！巨水！巨水！巨水！巨水！

var

l,n,k,i,j:longint;

s:char;

a:array[0..40]of longint;

d:array[0..41,0..41]of longint;

f:array[0..41,0..7]of longint;

begin

readln(n,k);

for i:=1to n do

begin

read(s);

a[i]:=ord(s)-ord('0');

d:=a[i];

end;

for i:=1to n do

for j:=i+1to n do

d:=d*10+a[j];

for i:=1to n do

f:=d[1,i];

for i:=1to k do

for j:=1to n do

for l:=i to j-1do

if f[l,i-1]*d[l+1,j]>f[j,i]

then f[j,i]:=f[l,i-1]*d[l+1,j];

writeln(f[n,i]);

end.

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

水题！水题！水题！水题！水题！水题！水题！水题！水题！

var

n,k,i,j,q,x:longint;

a:array[0..41]of longint;

f:array[-1..41,0..7]of longint;

s:string;

function shu(l,r:longint):longint;

var i:longint;

begin

shu:=0;

for i:=l to r do

shu:=shu*10+a[i];

end;

begin

readln(n,k);

k:=k+1;

readln(s);

filldword(f,sizeof(f) div 4,1);

for i:=1 to n do

a[i]:=ord(s[i])-48;

for i:=1 to n do

for j:=1 to k do

for q:=i-1 downto j-1 do

begin

x:=shu(q+1,i);

if x*f[q,j-1]>f

then f:=x*f[q,j-1];

end;

writeln(f[n,k]);

end.

数据。。。。无语！！

注意

这一题一定要用字符串读入！

program p1182;

type xxx=array[1..240] of integer;

var a,b,c,d,k,n,l:longint;

p,t,ans:xxx;

x:array[1..6,1..40] of xxx;

function chen(i,j:xxx;e:longint):xxx;

var a,l,b:longint;

t:xxx;

begin

l:=100;

while i[l]=0 do dec(l);

fillchar(t,sizeof(t),0);

for a:=1 to l do

for b:=1 to e do

begin

t[a+b-1]:=t[a+b-1]+i[a]*j;

t[a+b]:=t[a+b]+t[a+b-1] div 10;

t[a+b-1]:=t[a+b-1] mod 10;

end;

chen:=t;

end;

function max(i,j:xxx):xxx;

var a,b,c:longint;

begin

a:=100;

while (a=0) or (i[a]=j[a]) do

begin

dec(a);

end;

if i[a]>j[a] then max:=i

else max:=j;

end;

procedure init;

var s:string;

begin

readln(n,k);

readln(s);

for a:=1 to n do

p[a]:=ord(s[a])-48;

end;

procedure main(i,j:longint);

var a,b,c:longint;

s:xxx;

begin

for a:=1 to 240 do if x[a]0 then exit;

for a:=i-1 to j-1 do

begin

fillchar(s,sizeof(s),0);

for b:=a+1 to j do

s[j-b+1]:=p;

main(i-1,a);

x:=max(x,chen(x,s,j-a));

end;

end;

begin

init;

fillchar(x,sizeof(x),0);

for a:=1 to n-k do

for b:=1 to a do

x[1,a][a-b+1]:=p;

main(k+1,n);

a:=100;

while x[k+1][n][a]=0 do dec(a);

for b:=a downto 1 do write(x[k+1][n]);

writeln;

end.

一次AC

想了下,貌似数据有点小就写了integer,错了一次,日啊,我的ac率

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

哈哈，有了下面那只的前车之鉴，1次AC 哈哈哈。

原来是循环问题。无语……

预处理b数组

然后dp

for i:=2 to n do

for j:=1 to i-1 do

begin

max:=0;

for l:=1 to i-1 do

begin

if a[l,j-1]*b[l+1,i-l]>max then max:=a[l,j-1]*b[l+1,i-l];

end;

a:=max;

end;