P1087 FBI树

题解

 1 #include<iostream>
 2 #include<cmath>
 3 
 4 using namespace std;
 5 
 6 int n;
 7 char s[1050];
 8 
 9 void build_tree(int l,int r){
10     if(r>l){
11         int mid=(l+r)/2;
12         build_tree(l,mid);
13         build_tree(mid+1,r);
14     }
15     int ans=0,pos=0;
16     for(int i=0;i<=r-l;i++){
17         if(s[l+i]==‘1‘){
18             ans++;
19         }
20         else pos++;
21     }
22     if(pos==0) cout<<‘I‘;
23     else if(ans==0) cout<<‘B‘;
24     else cout<<‘F‘;
25 }
26 
27 
28 int main(){
29     cin>>n>>s;
30     int t=pow(2,n);        //构建的树的大小为 2^(n+1)-1; 
31     build_tree(0,t-1);
32     return 0;
33 }