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;
}

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