var
s:ansistring;
p,t,i,j,k,m,n:longint;
a:array[1..3000] of char;
ch:char;
procedure f(t:longint);
begin
if t>=p then
begin
write(a[t]);
exit;
end;
f(t*2);
f(t*2+1);
write(a[t]);
end;
begin
readln(n);
p:=2;
if n=0 then
begin
read(ch);
if ch='1' then write('I') else write('B');
halt;
end;
for i:=1 to n-1 do
p:=p*2;
for i:=1 to p do
begin
read(ch);
if ch='1' then
a[i+p-1]:='I' else
a[i+p-1]:='B';
end;
a[1]:=' ';
k:=p;
while a[1]=' ' do
begin
i:=k;
while i<2*k do
begin
if (a[i]='I') and (a[i+1]='I') then
a[i div 2]:='I' else
if (a[i]='B') and (a[i+1]='B') then
a[i div 2]:='B' else
a[i div 2]:='F';
i:=i+2;
end;
k:=k div 2;
end;
f(1);
end.
15ms...刚开始数组开小了。。。。。。。。唉、。。。。我的AC率。。。

#include<cstdio>
bool S[1025]={false};
char tmp;
int n,tbl[11]={1,2,4,8,16,32,64,128,256,512,1024};
char dfs(int s,int e){
if(s==e){
if(S[s]){printf("I");return 'I';}
else{printf("B");return 'B';}
}else{
char s1=dfs(s,(s+e)/2),s2=dfs((s+e)/2+1,e);
if(s1=='F'||s2=='F'){printf("F");return 'F';}
else if(s1=='B'&&s2=='B'){printf("B");return 'B';}
else if(s1=='I'&&s2=='I'){printf("I");return 'I';}
else{printf("F");return 'F';}
}
}
int main(){
//freopen("P1114.in","r",stdin);
//freopen("P1114.out","w",stdout);
scanf("%d\n",&n);
n=tbl[n];
for(int i=0;i<n;i++){
scanf("%c",&tmp);
if(tmp=='0')S[i]=false;
else S[i]=true;
}
dfs(0,n-1);
printf("\n");
}

#构造一棵二叉树（确切地说是满二叉树），后序遍历它

C：

#include <stdio.h>
#include <string.h>
#define MAXL 0x400/**最多1024位**/

void main(){
char s[MAXL];
int n;
void work(char b[]);

scanf("%d", &n);
scanf("%s", s);
work(s);/**分解字符串 s **/
}

void work(char base[])
{
int l;
char left[MAXL], right[MAXL];
char call(char a[]);/**判断输出‘F’ or 'B' or 'I' /
void stringcopy(char a[], char b[], int i, int j);/把字符串b的i~j位复制到a上**/

l = strlen(base);
if(l > 1){
stringcopy(base, left, 0, l/2);/**左孩子**/
stringcopy(base, right, l/2, l);/**右孩子**/
work(left);/**递归左孩子**/
work(right);/**递归右孩子**/
}
printf("%c", call(base));
}

char call(char base[])
{
int i, l;
l = strlen(base);
for(i = 0; i < l-1; i++)
if(base[i] != base[i+1]) return 'F';
if(base[i] == '0') return 'B';
if(base[i] == '1') return 'I';
}

void stringcopy(char base[], char get[], int st, int ed)
{
int i = st;
for(;i < ed; i++)
get[i-st] = base[i];
get[ed-st] = '\0';/**插入空字符，表示字符串的结束**/
}

var a:array[1..100000] of boolean;
n,m,i:longint;
b:char;
function aa(x,y:longint):longint;
var i,j,k,l,q,w,e:longint;
begin
if x=y then
begin
if a[x] then k:=1 else k:=0;
if k=1 then write('I') else write('B');
exit(k);
end;
k:=aa(x,(x+y) div 2);
l:=aa((x+y) div 2+1,y);
if (k=l)and(k=1) then begin e:=1;write('I');end else
if (k=l)and(k=0) then begin e:=0;write('B');end else
begin e:=2;write('F');end;
exit(e);
end;
begin
readln(n);m:=1;
for i:=1 to n do m:=m*2;n:=m;
for i:=1 to n do begin read(b);if b='1' then a[i]:=true else a[i]:=false;end;
i:=aa(1,n);
writeln;
end.

15ms
随随便便后序遍历

测试数据 #0: Accepted, time = 0 ms, mem = 440 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 440 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 440 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 432 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 440 KiB, score = 10

#include <stdio.h>
char fbi[3000];
int tep=0;
void set(char *s,int l,int r)
{
int i,flag=0;
int mid=(l+r)>>1;
if(l==r)
{
if(s[l]=='0')fbi[tep++]='B';
if(s[l]=='1')fbi[tep++]='I';
return ;
}
set(s,l,mid);
set(s,mid+1,r);
if(s[l]=='0')fbi[tep]='B';
if(s[l]=='1')fbi[tep]='I';
for(i=l+1;i<=r;i++)
{
if(s[l]!=s[i])flag=1;
}
if(flag)fbi[tep]='F';
tep++;
}
int main()
{
char s[2000];
int a,n,m,tep;
scanf("%d",&a);
n=1<<a;
scanf("%s",s);
set(s,0,n-1);
m=n*2-1;
fbi[m]='\0';
printf("%s",fbi);
return 0;
}

15MS呢。。。那些人怎么做到100多ms的。。
测试数据 #0: Accepted, time = 15 ms, mem = 548 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 464 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 468 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 472 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 508 KiB, score = 10
Accepted, time = 15 ms, mem = 548 KiB, score = 100

#include <iostream>
#include <cmath>
#include <string>
using namespace std;

struct TreeNode {
char type;
TreeNode *l, *r;
} *root;

char FBI[1030];

int n;

string substring(const string &str, int l, int r) {
return str.substr(l, r-l+1);
}

void create(TreeNode *&root, int l, int r) {
root = new TreeNode;
if(r-l+1==1) {
root->l = root->r = NULL;
if(FBI[l]=='1') {
root->type = 'I';
} else root->type = 'B';
}
else {
int m = (l+r)/2;
create(root->l, l, m);
create(root->r, m+1, r);
if(root->l->type==root->r->type&&root->l->type!='F') {
root->type = root->l->type;
} else root->type = 'F';
}
}

void last_order(TreeNode *r) {
if(!r) return;
last_order(r->l);
last_order(r->r);
cout<<r->type;
}

int main() {
cin>>n;
cin>>FBI + 1;
n = pow(2, n);
create(root, 1, n);
last_order(root);
return 0;
}

辛苦写完的程序第一次测评时竟被测评机忽略了………………
包子，你的代码的变量名称也太…………有内涵了，表示正常人看不懂啊
给个改良版
测试数据 #0: Accepted, time = 78 ms, mem = 780 KiB, score = 10

测试数据 #1: Accepted, time = 62 ms, mem = 520 KiB, score = 10

测试数据 #2: Accepted, time = 46 ms, mem = 520 KiB, score = 10

测试数据 #3: Accepted, time = 62 ms, mem = 520 KiB, score = 10

测试数据 #4: Accepted, time = 62 ms, mem = 516 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 520 KiB, score = 10

测试数据 #6: Accepted, time = 15 ms, mem = 520 KiB, score = 10

测试数据 #7: Accepted, time = 15 ms, mem = 520 KiB, score = 10

测试数据 #8: Accepted, time = 31 ms, mem = 520 KiB, score = 10

测试数据 #9: Accepted, time = 46 ms, mem = 784 KiB, score = 10

Summary: Accepted, time = 432 ms, mem = 784 KiB, score = 100
代码

program FBI;
type
rec=record
data:char;
lch,rch,p:integer;
end;
var
n,sum,i,j:integer;
f:array[1..5000]of rec;
s:ansistring;
a:array[1..5000]of integer;
function word(x,y:integer):char;
begin
word:='F';
if (f[x].data='B')and(f[y].data='B') then
word:='B'
else if (f[x].data='I')and(f[y].data='I') then
word:='I';
end;
procedure build1(i:integer);
begin
if (i>sum) then
exit
else begin
f[i].p:=i div 2;
f[i].lch:=2*i;
f[i].rch:=2*i+1;
inc(i);
build1(i);
end;
end;
procedure behind(i:integer);
begin
if f[i].data='#' then
exit
else begin
if (f[i].lch>0)and(f[i].lch<2*sum) then
behind(f[i].lch);
if (f[i].rch>0)and(f[i].rch<2*sum) then
behind(f[i].rch);
write(f[i].data);
end;
end;
begin
readln(n);
sum:=1;
for i:=1 to n do
sum:=sum*2;
read(s);
build1(1);
for i:=1 to sum do
a[i]:=ord(s[i])-ord('0');
for i:=1 to 2*sum+1 do
f[i].data:='#';
for i:=sum to 2*sum-1 do
begin
f[i].data:='F';
if a[i-sum+1]=0 then
f[i].data:='B'
else if a[i-sum+1]=1 then
f[i].data:='I';
end;
for i:=sum-1 downto 1 do
f[i].data:=word(f[i].lch,f[i].rch);
behind(1);
end.

106ms无压力秒杀~
type nn=record
f,l,r:longint;
n:char;
end;

var
i,n,t,l,top:longint;
s:ansistring;
a:array[1..10000]of longint;
f:array[1..10000]of nn;

procedure lala(i:longint);
begin
f[i].f:=i div 2;
f[i].l:=i*2;
f[i].r:=i*2+1;
i:=i+1;
if i>l then exit;
lala(i);
end;

procedure ee(i:longint);
begin
if (f[i].l<>0)and(f[i].l<l*2) then ee(f[i].l);
if (f[i].r<>0)and(f[i].r<l*2) then ee(f[i].r);
write(f[i].n);
end;

Begin
readln(n);
l:=1;
for i:=1 to n do l:=l*2;
readln(s);
for i:=1 to l do a[i]:=ord(s[i])-ord('0');
top:=2*l-1;
lala(1);
for i:=l to top do if a[i-l+1]=0 then f[i].n:='B' else f[i].n:='I';
for i:=l-1 downto 1 do begin
f[i].n:='F';
if (f[f[i].l].n='I') and (f[f[i].r].n='I') then f[i].n:='I';
if (f[f[i].l].n='B') and (f[f[i].r].n='B') then f[i].n:='B';
end;
ee(1);
End.

├ 测试数据 01：答案正确... (0ms, 1032KB)

├ 测试数据 02：答案正确... (0ms, 664KB)

├ 测试数据 03：答案正确... (0ms, 628KB)

├ 测试数据 04：答案正确... (0ms, 664KB)

├ 测试数据 05：答案正确... (0ms, 664KB)

├ 测试数据 06：答案正确... (0ms, 664KB)

├ 测试数据 07：答案正确... (0ms, 664KB)

├ 测试数据 08：答案正确... (0ms, 664KB)

├ 测试数据 09：答案正确... (0ms, 664KB)

├ 测试数据 10：答案正确... (0ms, 1032KB)

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

Accepted / 100 / 0ms / 1032KB

思路：按题目要求建造满二叉树tree，再后序遍历输出

{

ID:darkgod-z

PROG:vijos P1114

HANG:PASCAL

}

var

s:ansistring;

n:integer;

tree:array[1..2047] of record

data:char;

l,r:integer;

end;

procedure lrd(x:integer);

begin

if x0 then begin

lrd(tree[x].l);

lrd(tree[x].r);

write(tree[x].data);

end;

end;

procedure gettree(s:ansistring; i:integer);

var

s1,s2:ansistring;

t,j:integer;

begin

if (pos('0',s)>0)and(pos('1',s)>0) then tree[i].data:='F'

else

if pos('0',s)>0 then tree[i].data:='B'

else tree[i].data:='I';

if length(s)=1 then exit;

t:=length(s) div 2;

for j:=1 to t do s1:=s1+s[j];

for j:=t+1 to length(s) do s2:=s2+s[j];

if i*2

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

Ansistring

第一次做到了一次AC，祝贺一下，用的递归

#include

int cifang[12]={1,2,4,8,16,32,64,128,256,512,1024,2048};

int panduan(char* s,int n)

{

int i;

char c=s[0];

int flag=1;

for(i=1;i

├ 测试数据 01：运行时错误... (83ms, 1032KB)

读取访问违规, 地址: 0x0000d7aa

├ 测试数据 02：答案正确... (64ms, 700KB)

├ 测试数据 03：答案正确... (40ms, 628KB)

├ 测试数据 04：答案正确... (118ms, 664KB)

├ 测试数据 05：答案正确... (142ms, 664KB)

├ 测试数据 06：答案正确... (103ms, 664KB)

├ 测试数据 07：答案正确... (173ms, 700KB)

├ 测试数据 08：答案正确... (169ms, 700KB)

├ 测试数据 09：答案正确... (173ms, 700KB)

├ 测试数据 10：答案正确... (185ms, 996KB)

————————————

谁能告诉我第一个点是怎么个思想？

————已经解决了 是数组开小了 2的10次方是1024 我开1025 挂了， 开3000 ac！

求后序遍历符合递归的规律(都是树的结构),

十分同意楼下所说的,关键在于理解题意

做法:

将题目所给的树不断按中间将树切开两半,拼命切,

直到只有一个结点为止,然后返回到自己的父亲,拼命返回,直到根结点为止

用递归实现

关键在于理解题意。。

var

f:array[1..2000] of char;

n,i,j,k,tt:integer;

procedure search(s,t:integer);

var

tot1,tot0:integer;

p:char;

begin

if t-s+11 then

begin

tot1:=0; tot0:=0;

for i:=s to t do

if f[i]='1' then inc(tot1) else inc(tot0);

if tot1=0 then p:='B' else

if tot0=0 then p:='I' else

p:='F';

search(s,(t-s+1) div 2+s-1);

search((t-s+1) div 2+s,t);

write(p);

end;

end;

begin

readln(n);

tt:=1;

for i:=1 to n do tt:=tt*2;

for i:=1 to tt do read(f[i]);

search(1,tt);

writeln;

end.

我的代码应该比较简洁，清一色的分治。

#include

using namespace std;

int n;

char a[1025];

void init(void)

{

cin>>n>>a;

return;

}

int fbi(int i,int j)

{

if(i0)cout

晕 建造树居然忘了 New(p)！！！

program fbi;

const

fbii:array[0..2]of char=('B','I','F');

var

n,i,t:longint;

shu:ansistring;

shui:array[1..1024]of longint;

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

var x,y:longint;

begin

if l=r then

begin

if shui[l]=1 then begin tree:=1;write(fbii[1]);end;

if shui[l]=0 then begin tree:=0;write(fbii[0]);end;

end else

begin

x:=tree(l,(l+r) div 2);

y:=tree((l+r) div 2+1,r);

if x+y=0 then begin tree:=0;write(fbii[0]);end

else

if x*y=1 then begin tree:=1;write(fbii[1]);end

else begin tree:=2;write(fbii[2]);end

end;

end;

begin

readln(n);

readln(shu);

t:=length(shu);

for i:=1 to t do

shui[i]:=ord(shu[i])-48;

tree(1,t);

end.

好水好喝！

WATER...

题目让我终于看懂了。。。

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

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

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

Accepted 有效得分：100 有效耗时：0ms

program ex;

var i,j,n:longint;

st:ansistring;

procedure get(l,r:longint);

var i,j:longint;

begin

if r-l>=1 then

begin

get(l,(l+r)div 2);

get((l+r)div 2+1,r);

end;

if pos('0',copy(st,l,r-l+1))=0 then

write('I')

else

if pos('1',copy(st,l,r-l+1))=0 then

write('B')

else

write('F');

end;

begin

readln(n);

readln(st);

get(1,length(st));

writeln;

end.