2010-09-21北大创新工场笔试大题

题目：

有两个量杯A和B，分别可以量M和N升水，还有装满水的池子，以及一个大的空杯子C，现在要经过一系列的步骤在空杯子里面量出K升水。编写程序，输入为M，N，K，输出为步骤，或者“No Solution”。例如M, N, K分别为5, 7, 4，则输出的步骤为：

0, 0, 0         (都没有水)

 

0, 7, 0         (B倒满水)

5, 2, 0         (B倒入A)

5, 0, 2         (B到入C)

0, 0, 2         (B倒入水池)

0, 7, 2         (B倒满水)

 

5, 2, 2         (B倒入A)

 

5, 0, 4         (B到入C，达到要求)

 

采用回溯法

#include <iostream>#include <stack>#include <cstdlib>using namespace std;enum Action{ACTION_A,ACTION_B,ACTION_AB,ACTION_A2B,};static void print_action(const int M, const int N, int ∑,const Action a, const bool a_larger){switch (a){case ACTION_A:cout << M << ", 0, " << sum << endl; sum += M;cout << 0 << ", 0, " << sum << endl; break;case ACTION_B:cout << "0, " << N << ", " << sum << endl; sum += N;cout << "0, 0, " << sum << endl; break;case ACTION_AB:if (a_larger){cout << M << ", 0, " << sum << endl; cout << M - N << ", " << N << ", " << sum << endl; sum += (M - N);cout << "0, " << N << ", " << sum << endl; }else{cout << "0, " << N << ", " << sum << endl; cout << M << ", " << N - M << ", " << sum << endl; sum += (N - M);cout << M << ", " << 0 << ", " << sum << endl; }cout << "0, 0, " << sum << endl;break;case ACTION_A2B:break;default:cout << "error in print_action" << endl;exit(-1);}}static void print_solution(const int M, const int N,stack<Action> &action, const bool a_larger){Action a;int sum = 0;while (!action.empty()){a = action.top();print_action(M, N, sum, a, a_larger);action.pop();}}static void try_to_fill(const int M, const int N, const int K, stack<Action> &action, const bool a_larger){if (0 == K){print_solution(M, N, action, a_larger);exit(0);}else if (K < 0){return;}action.push(ACTION_A);try_to_fill(M, N, K - M, action, a_larger);action.pop();action.push(ACTION_B);try_to_fill(M, N, K - N, action, a_larger);action.pop();action.push(ACTION_AB);try_to_fill(M, N, K - (a_larger ? (M - N) : (N - M)), action, a_larger);action.pop();}static void simple_fill(const int M, const int K, stack<Action> &action){int tmp = K;while (tmp > 0){action.push(ACTION_A);tmp -= M;}if (0 == tmp){print_solution(M, M, action, 0);}else{cout << "No Solution" << endl;}}int main(){const int M = 5, N = 7, K = 4;stack<Action> action;bool a_larger = (M > N);cout << M << ", " << N << ", " << K << endl;cout << "******************" << endl;cout << "0, 0, 0" << endl; /*all empty*/if (M == N){simple_fill(M, K, action);return 0;}try_to_fill(M, N, K, action, a_larger);action.push(ACTION_A2B);try_to_fill(M, N, K - (a_larger ? (2*N - M) : (2*M - N)), action, a_larger);cout << "No Solution" << endl;return 0;}