参考了解答,用枚举遍历的方式实现。
#include <bits/stdc++.h> using namespace std; double cal(double a, double b, char ch) { switch (ch) { case '+': return a + b; case '-': return a - b; case '*': return a * b; case '/': return a / b; } return 0; } bool check(vector<int> &nums) { char op[4] = {'+','-','*','/'}; sort(nums.begin(), nums.end()); do { // 遍历三个位置所有可能的 for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { for (int k = 0; k < 4; k++) { double first = cal(nums[0], nums[1], op[i]); double second = cal(first, nums[2], op[j]); if (cal(second, nums[3], op[k]) == 24) return true; } } } } while (next_permutation(nums.begin(), nums.end())); return false; } int main() { vector<int> nums; for (int i = 0; i < 4; i++) { int temp = 0; cin >> temp; nums.push_back(temp); } if (check(nums)) { cout << "true" << endl; } else { cout << "false" << endl; } return 0; }