用枚举类加上了线段树的建树思想，醉了。。
C++
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
enum treeType {B,I,F};
ostream & operator<<(ostream & out, const treeType & display)
{
switch (display)
{
case B:out << 'B'; break;
case I:out << 'I'; break;
case F:out << 'F'; break;
default:break;
}
return out;
}
treeType operator+(treeType & a, treeType & b)
{
treeType ans;
if (a == b) ans = a;
else ans = F;
return ans;
}
template <typename T>
void LRD(vector<T> &tree,int root=1)
{
if (root < tree.size())
{
LRD(tree, root * 2);
LRD(tree, root * 2 + 1);
cout << tree[root];
}
}
int main()
{
int n;
cin >> n;
int numOfnode = pow(2, n+1)-1;
int leafBegin = pow(2, n );
// cout << numOfnode << endl << leafBegin << endl;
vector<treeType> tree(numOfnode+1);
char temp;
for (int i = leafBegin; i <= numOfnode; i++)
{
cin >> temp;
if (temp == '0') tree[i] = B;
else tree[i] = I;
}
for (int i = leafBegin - 1; i >= 1; i--)
{
tree[i] = tree[2 * i] + tree[2 * i + 1];
}
LRD(tree);
//system("pause");
return 0;
}


var n:longint;
  s:ansistring;
procedure fbi(s:ansistring);
var len,a,b:longint;
  begin
    len:=length(s);
    if len=1 then if s[1]='0' then write('B') else write('I')
    else
      begin
        fbi(copy(s,1,len div 2));
        fbi(copy(s,len div 2+1,len-len div 2));
        a:=pos('0',s);
        b:=pos('1',s);
        if (a<>0) and (b<>0) then write('F');
        if (a=0) and (b<>0) then write('I');
        if (a<>0) and (b=0) then write('B');
      end;
  end;
begin
  readln(n);
  readln(s);
  fbi(s);
  readln;
  end.

楼下好6，我的15ms

水题不多说，**以空间换时间**
0ms与1240kb的苦逼程序
直接分治~
注意**string过不了要ansistring！！**
```pascal
program FBI;
var bi:array[1..10000]of char;
n,i,count:longint;
s:ansistring;

function judge(a:ansistring):char;
var i:longint;
t:char;
begin
t:=a[1];
for i:=2 to length(a) do
if t<>a[i] then exit('F');
if t='0' then exit('B');
exit('I');
end;

procedure merge(s:ansistring;n:longint);
var i:longint;
begin
if n=1 then
begin
if s='0' then write('B') else write('I');
exit;
end;
merge(copy(s,1,length(s)div 2),n div 2);
merge(copy(s,length(s)div 2 +1,length(s)),n div 2);
write(judge(s));
end;

begin
readln(n);
readln(s);
count:=1;
for i:=1 to n do count:=count*2;
merge(s,count);
end.

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <utility>
#include <map>
#include <set>
#include <climits>
#include <stack>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

struct TreeNode
{
    char value;
    TreeNode* left_child;
    TreeNode* right_child;
};

char str[1030];
int n;

TreeNode* BuildTree(char* str, int len)
{
    if (len == 0)
        return 0;

    int type = str[0] - '0';
    if (len != 1)
    {
        for (int i = 1; i < len; i++)
        {
            if (str[i] != str[0])
            {
                type = -1;
                break;
            }
        }
    }

    TreeNode* node = new TreeNode;

    if (type == -1)
    {
        node->value = 'F';
    }
    else if (type == 0)
    {
        node->value = 'B';
    }
    else
    {
        node->value = 'I';
    }

    if (len == 1)
    {
        node->left_child = 0;
        node->right_child = 0;
    }
    else
    {
        node->left_child = BuildTree(str, len / 2);
        node->right_child = BuildTree(str + len / 2, len - len / 2);
    }

    return node;
}

void PostOrder(TreeNode* node)
{
    if (node == 0)
        return ;
    PostOrder(node->left_child);
    PostOrder(node->right_child);
    printf("%c", node->value);
}

int main()
{
    while(~scanf("%d", &n))
    {
        getchar();
        gets(str);
        //puts(str);
        TreeNode* node = BuildTree(str, strlen(str));
        //cout << "build success" << endl;
        PostOrder(node);
        printf("\n");
    }

    return 0;
}

#include <iostream>
#include <cstring> 
#include <cstdio>
using namespace std;
int n;
char a[1025];
inline void fbi(int i,int j){
    if(i<=j){
        int mid=(i+j)/2,I=0,B=0;
        if(i!=j){
        fbi(i,mid);
        fbi(mid+1,j);
        }
        while(i<=j)if(a[i++]=='0')B++;else I++;
        if(B>0&&I>0)cout<<'F';
        else if(B>0)cout<<'B';
        else cout<<'I';
    }
}
int main(int argc, char const *argv[]){
    /* code */
    //freopen("155.in","r",stdin);
    //freopen("155.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>a;
    fbi(0,strlen(a)-1);
    return 0;
}

发一个弱弱的题解，思路是模拟栈，并在栈里面增设一个flag数组，遇到栈中超过两个字符并且两个字符的flag相同就进行处理成一个字符并且这个字符的flag+1；

include <stdio.h>

include <math.h>

char a[1500][2];

void puttree(char c)
{
if (c == '1')
putchar('I');
else if (c == '0')
putchar('B');
else
putchar('F');
}

int stack_push(int top)
{
if (a[top][0] != a[top-1][0]) {
a[top-1][0] = '2';
}
a[top-1][1]++;
puttree(a[top-1][0]);
return top-1;
}

int main (void)
{
int top = -1, n, flag, i, power;
char c;
scanf("%d",&n);
getchar();
power = pow(2,n);
for (i=0; i<power; i++) {
c = getchar();
a[++top][0] = c;
a[top][1] = flag;
puttree(c);
while (top >= 1 && a[top][1] == a[top-1][1]) {
top = stack_push(top);
}
}
putchar('\n');
return 0;
}

水题
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100003;
int n, t[1025];
char s[1025];
void dfs(int l, int r) {
if (l == r) {putchar(t[l] == 1 ? 'I' : 'B'); return;}
int mid = (l + r) >> 1;
dfs(l, mid); dfs(mid + 1, r);
bool flag0 = 0, flag1 = 0;
for(int i = l; i <= r; ++i)
if (t[i] == 0) flag0 = 1;
else flag1 = 1;
if (flag0 && flag1) putchar('F');
else putchar(flag0 ? 'B' : 'I');
}
int main() {
scanf("%d", &n);
scanf("%s", s);
int len = strlen(s);
for(int i = 0 ;i < len; ++i)
t[i + 1]=s[i]-'0';
dfs(1, len);
return 0;
}

var
i,n:longint;
a:array[1..1024] of char;

procedure try(p,q:longint);
var
n0,n1,i:longint;
begin
n0:=0;
n1:=0;
for i:=p to q do
if a[i]='0' then inc(n0)
else inc(n1);
if p=q then
begin
if n1=1 then begin write('I'); exit; end;
if n0=1 then begin write('B'); exit; end;
end;
try(p,(p+q) div 2);
try((p+q) div 2+1,q);
if q-p+1=n0 then write('B')
else if q-p+1=n1 then write('I')
else write('F');
end;

begin
readln(n); n:=1 shl n;
for i:=1 to n do read(a[i]);
try(1,n);
end.

** 题解与代码**

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 5000 + 24;
inline int LC(int x) {return x << 1;}
inline int RC(int x) {return x << 1 | 1;}
int n;
char s[N];
char sum[N << 2];
void Update(int x)
{
    if(sum[LC(x)] == sum[RC(x)])
        sum[x] = sum[LC(x)];
    else
        sum[x] = 'F';
    printf("%c",sum[x]);
}
void Buile(int x, int l, int r)
{
    if(l == r)
    {

        s[l-1] == '0' ? sum[x] = 'B' : sum[x] = 'I';
        printf("%c",sum[x]);
        return;
    }
    int mid =(l + r) >> 1;
    Buile(LC(x),l, mid);
    Buile(RC(x),mid+1,r);
    Update(x);
}
int main()
{
    while(~scanf("%d",&n))
    {
        scanf("%s",&s);
        Buile(1,1,1 << n);
        printf("\n");
    }
}

不知道为什么出错啊

import java.io.IOException;
import java.util.Scanner;

public class Main {

static char a[] = new char[1100];
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);

int n = sc.nextInt();
int sum = 1;
for(int i=0;i<n;i++)
sum = sum*2;
for(int i=0;i<sum;i++){
a[i] = (char)System.in.read();
}
judge(0,sum-1);
}
private static char judge(int a1,int a2){
int i,t0,t1;
if(a1 == a2){
if(a[a1] == '0') {System.out.print("B");return 'B';}
else System.out.print("I");return 'I';
}
else{
i = (a2-a1-1)/2;
char j1 = judge(a1,a1+i);
char j2 = judge(a2-i,a2);
t0=0; t1=0;
if(j1 == 'I'&&j2 == 'I') {System.out.print("I");return 'I';}
else if(j1 == 'B'&&j2 == 'B') {System.out.print("B");return 'B';}
else System.out.print("F");return 'F';
}
}
}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 5000 + 24;
inline int LC(int x) {return x << 1;}
inline int RC(int x) {return x << 1 | 1;}
int n;
char s[N];
char sum[N << 2];
void Update(int x)
{
if(sum[LC(x)] == sum[RC(x)])
sum[x] = sum[LC(x)];
else
sum[x] = 'F';
printf("%c",sum[x]);
}
void Buile(int x, int l, int r)
{
if(l == r)
{

s[l-1] == '0' ? sum[x] = 'B' : sum[x] = 'I';
printf("%c",sum[x]);
return;
}
int mid =(l + r) >> 1;
Buile(LC(x),l, mid);
Buile(RC(x),mid+1,r);
Update(x);
}
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",&s);
Buile(1,1,1 << n);
printf("\n");
}
}

构造完全二叉树即可。
就是不知道为什么在本地测试时输出到文件总是有问题。
问下：cin不是会过滤行末的空格么，为什么cin>>n后还要加一句getchar()........
/*
ID:
LANG:
TESK:
*/
#include <iostream>
#include <fstream>
//#include <cstdio>
//#include <cstdlib>
#include <string>
//#include <cstring>
//#include <cmath>
//#include <algorithm>
using namespace std;
typedef long long longint;
const string file_name = "p1114";
const string file_in_name = file_name+".in";
const string file_out_name = file_name+".out";
const int N=2048;
//const int M=100;
longint n,m;
string st;
string t[N];
char f[N];
char cha(string s)
{
longint i=1,l=s.length();
while(s[i]==s[i-1]&&i<l)i++;
if(i==l)
if(s[0]=='1')return 'I';
else return 'B';
else return 'F';
}
int pro(longint i)
{
longint l=t[i].length();
if(l<=1)return 0;
l=l>>1;
longint l1=i<<1,l2=l1+1;
t[l1].assign(t[i],0,l);
f[l1]=cha(t[l1]);
t[l2].assign(t[i],l,l);
f[l2]=cha(t[l2]);
pro(l1);
pro(l2);
return l;
}
int print(longint i)
{
if(i>=(1<<n)&&i<(1<<(n+1))){cout<<f[i];return 0;}
print(i<<1);
print((i<<1)+1);
cout<<f[i];
return 0;
}
int main()
{
//ifstream cin(file_in_name.c_str(),ios::in);
//ofstream cout(file_out_name.c_str(),ios::out);
//freopen(file_in_name.c_str(),"r",stdin);
//freopen(file_out_name.c_str(),"w",stdout);
cin>>n;
getchar();
getline(cin,t[1]);
f[1]=cha(t[1]);
//cout<<t[1];
//cout<<cha(t[1]);
pro(1);
//for(int i=1;i<(1<<(n+1));i++)cout<<i<<':'<<f[i]<<endl;
print(1);
//cout<<f[1];
//cin.close();
//cout.close();
return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

int num[1030];
char k[1030];

void check(int a,int b){
int c0 = 0, c1= 0;
for(int i=a; i<=b; i++){
if(num[i] == 1)
c0 = 1;
if(num[i] == 0)
c1 = 1;
if(c0 == 1 && c1 == 1){
cout<<"F";
return;
}
}
if(c0 == 1)
cout<<"I";
else
cout<<"B";
}

void back(int a, int b){
if(a == b){
check(a, b);
return;
}
if(a > b)
return;
back(a, (a+b)/2);
back((a+b)/2+1, b);
check(a, b);
}

int main()
{
int n;
cin>>n;
int bri=1;
for(int i=1; i<=n; i++)
bri *= 2;
for(int i=1; i<=bri; i++){
cin>>k[i];
num[i] = k[i] - '0';

}
back(1, bri);
system("pause");
return 0;
}
水一发。。。

完全不懂树结构，只好直接按题目模拟过程

#include<stdio.h>
int main( )
{
char a[12][1500],t;
int i,n,l=1,j,k,flag[11]={0},mol[11]={0};

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

for(i=1;i<=n;i++)
l=l*2;

for(i=1;i<=l;i++)
{
scanf("%c",&t);
if (t=='0') a[1][i]='B';
else a[1][i]='I';
}

k=l;

for(i=2;i<=n+1;i++)
{

for(j=2;j<=k;j=j+2)
{
if(a[i-1][j]!=a[i-1][j-1]) a[i][j/2]='F';
else a[i][j/2]=a[i-1][j];
}

k=k/2;

}

for(i=1;i<=l;i++)
{
printf("%c",a[1][i]);
flag[1]++;

for(j=1;j<=n+1;j++)
if(flag[j]%2==0 && flag[j]!=mol[j]) {printf("%c",a[j+1][flag[j]/2]);mol[j]=flag[j];flag[j+1]++;}
}

return 0;

}

var
s:ansistring;
n,sq:longint;

procedure wtf(str:string);
var
i,j,k,m:longint;
s1,s2:ansistring;
begin
j:=0; k:=0; m:=length(str) div 2;
if length(str)=1 then
begin
if str='0' then write('B')
else write('I');
end;
if (length(str)>1) then
begin
s1:=copy(str,1,m);
s2:=copy(str,m+1,m);
wtf(s1); wtf(s2);
for i:=1 to length(str) do
begin
if str[i]='0' then inc(j)
else inc(k);
end;
if (j=0) then write('I');
if (k=0) then write('B');
if (j<>0) and (k<>0) then write('F');
end;
end;

begin
readln(sq);
readln(s);
if (sq=0) then write('B')
else wtf(s);
readln();
end.

各位大神，请问为什么一直是80分，第一个和最后一个数据过不去

测试数据 #0: Accepted, time = 15 ms, mem = 196764 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 196504 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 196504 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 196500 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 196504 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 196508 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 196500 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 196500 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 196504 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 196760 KiB, score = 10
Accepted, time = 30 ms, mem = 196764 KiB, score = 100
代码
var
x,y,k,i,j,n,l,t,nv,nan,m:longint;
b,c,a:array[0..20001]of longint;
f:array[0..10000,0..10000]of integer;
s:ansistring;
procedure shuchu(s,w:longint);
var
i,j,k:longint;
begin
if s<w then shuchu(s,(s+w)div 2);
if s<w then shuchu((s+w)div 2+1,w);
case f[s,w] of
0:write('B');
1:write('I');
2:write('F');
end;
end;

procedure fbi(s,w:longint);
var
j,l,k:longint;
begin
if s=w then exit;
fbi(s,(s+w)div 2);
fbi((s+w)div 2+1,w);
if f[s,(s+w)div 2]=f[(s+w)div 2+1,w] then f[s,w]:=f[s,(s+w)div 2] else f[s,w]:=2; // writeln(f[s,w]);
end;

begin

readln(n); m:=1;
for i:=1 to n do m:=m*2;
readln(s);
for i:=1 to m do f[i,i]:=ord(s[i])-48;
//for i:=1 to m do write(f[i,i]);
readln;

fbi(1,m);
shuchu(1,m);

writeln;
readln;
end.

不用后序遍历，只要按照后序遍历的顺序构建就行了，根结点直接输出。
还有，string不够长，要用ansistring。

Pascal Code

var
s:ansistring;
n:longint;
function fbi(s:ansistring;len:longint):char;
var
i:longint;
flag0,flag1:boolean;
begin
flag0:=false;
flag1:=false;
for i:=1 to len do
begin
if s[i]='0' then flag0:=true;
if s[i]='1' then flag1:=true;
if flag0 and flag1 then exit('F');
end;
if flag0 then exit('B');
if flag1 then exit('I');
end;
procedure createtree(s:ansistring;len:longint);
begin
if len>1 then createtree(copy(s,1,len div 2),len div 2);
if len>1 then createtree(copy(s,len div 2+1,len div 2),len div 2);
write(fbi(s,len));
end;
begin
readln(n);
readln(s);
createtree(s,length(s));
end.

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char p[15000];
char x[15000];
char f(char ,int);
int main()
{ memset(x,'#',sizeof(x));
//freopen("fbi.in","r",stdin);
//freopen("fbi.out","w",stdout);
int n;
cin>>n;
scanf("%s",p);
void create(char,int,int);
create(p,1,strlen(p));
void h(int);
h(1);
return 0;

}

void h(int m)
{
if(x[m*2]!='#') h(m*2);
if(x[m*2+1]!='#') h(m*2+1);
cout<<x[m];
}

void create(char *s,int i,int len)
{

x[i]=f(s,len);
if(len>1)
{
create(s,i*2,len/2);
create(s+len/2,i*2+1,len/2);
}
}

char f(char *s,int len)
{ bool t1=false;
bool t2=false;
for(char *i=s;i!=s+len;++i)
{
if(*i=='0') t1=true;
if(*i=='1') t2=true;
}
if(t1==true&&t2!=true) return 'B';
if(t1!=true&&t2==true) return 'I';
if(t1==true&&t2==true) return 'F';
}

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.