Stacks of Flapjacks UVA

比较简单的一道题目，由大到小找出需要调整的数字的位置，然后将该数字翻转到最上方，然后再从目标位置进行一次翻转，将这数字调整到正确的位置，同时注意，如果该数字已经是在最上方了，那么就直接在目标位置处进行翻转就行了，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;string s;int amount,data[35];void flip(int ind){for (int i = 0; i < ind - i; i++){swap(data[i], data[ind - i]);}cout <<amount-ind << " ";}int main(){while (getline(cin, s)){if (s == "0") break;vector<int> res;stringstream is(s);int t;amount = 0;while (is >> t){    data[amount] = t;amount++;}cout << s << endl;for (int i = amount - 1; i >= 0; i--){int ind = max_element(data, data + i + 1) - data;if (ind == i) continue;if (ind > 0) flip(ind);flip(i);}cout << "0\n";}return 0;}