我我哦我我居然写出了递归

当时的想法就是每次对他进行对半分, 直到分到剩一个字符时, 到底

在这之前重复对半分, 递归, 现在献上代码

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 void FBI(string in, int len) {
 5     if (len > 1) {
 6         string in1, in2;
 7         in1 = in.substr(0, in.size() / 2);
 8         in2 = in.substr(in.size() / 2, in.size() / 2);
 9         FBI(in1, in1.size());
10         FBI(in2, in2.size());
11     }
12
13     if (in.find("1") != in.npos && in.find("0") != in.npos) cout << ‘F‘;
14     else if (in.find("1") != in.npos) cout << ‘I‘;
15     else cout << ‘B‘;
16 }
17
18 int main(void) {
19     int n;
20     string in;
21     cin >> n >> in;
22     FBI(in, in.size());
23
24     return 0;
25 }

原文地址：https://www.cnblogs.com/zhangzixu/p/11561116.html