SOJ 1050. Numbers & Letters

標籤：

題目大意：給出5個整數和4種運算(加法，減法，乘法和除法)，選取其中的部分數作任意運算，使得值等於目標數或者最接近目標數。其中，除法只能整除，即商必須是整數。

解題思路：每次選取任意兩個數，進行可行的運算，獲得結果並且和目標數作比較。最後得出答案。

代碼如下：

  1 #include <iostream>  2 //#include <ctime>  3 #include <vector>  4 #include <climits>  5 using namespace std;  6   7 const int maxn = 5;  8 int target;  9 int approximation; 10 int n; 11  12 bool test(const int &res) { 13     bool flag = false;; 14     if (res == target) { 15         approximation = res; 16         flag = true; 17     } else if (res < target) { 18         approximation = approximation > res ? approximation : res; 19     } 20     return flag; 21 } 22  23 int add(const int & a, const int & b) { 24     return a + b; 25 } 26  27 int sub(const int & a, const int & b) { 28     return a - b; 29 } 30  31 int mul(const int & a, const int & b) { 32     return a * b; 33 } 34  35 int div(const int & a, const int & b, bool &flag) { 36     int res = 0; 37  38     if (a < b) { 39         if (a != 0 && b % a == 0) { 40             res = b / a; 41             flag = true; 42         } else { 43             flag = false; 44         } 45     } else { 46         if (b != 0 && a % b == 0) { 47             res = a / b; 48             flag = true; 49         } else { 50             flag = false; 51         } 52     } 53  54     return res; 55 } 56  57 void deal(const vector<int> & nums) { 58     if (approximation == target) return; 59     if (nums.size() < 2) return; 60  61     for (int i = 0; i < nums.size(); i++) { 62         for (int j = i + 1; j < nums.size(); j++) { 63             vector<int> nnums; 64             nnums.insert(nnums.end(), nums.begin(), nums.begin() + i); 65             nnums.insert(nnums.end(), nums.begin() + i + 1, nums.begin() + j); 66             nnums.insert(nnums.end(), nums.begin() + j + 1, nums.end()); 67  68             int res; 69             res = add(nums[i], nums[j]); 70             if (test(res)) return; 71             nnums.push_back(res); 72             deal(nnums); 73  74             nnums.pop_back(); 75  76             res = sub(nums[i], nums[j]); 77             if (test(res)) return; 78             nnums.push_back(res); 79             deal(nnums); 80  81             nnums.pop_back(); 82  83             res = sub(nums[j], nums[i]); 84             if (test(res)) return; 85             nnums.push_back(res); 86             deal(nnums); 87  88             nnums.pop_back(); 89  90             res = mul(nums[i], nums[j]); 91             if (test(res)) return; 92             nnums.push_back(res); 93             deal(nnums); 94  95             nnums.pop_back(); 96  97             bool flag; 98             res = div(nums[i], nums[j], flag); 99             if (flag == false) continue;100             if (test(res)) return;101             nnums.push_back(res);102             deal(nnums);103 104             nnums.pop_back();105 106 107         }108     }109 }110 111 int main() {112     cin >> n;113     while (n--) {114         vector<int> nums;115         int temp;116         for (int i = 0; i < maxn; i++) {117             cin >> temp;118             nums.push_back(temp);119         }120         cin >> target;121         approximation = -INT_MAX;122 123         //double start = clock();124         deal(nums);125         //double end = clock();126         //cout << (end - start) / CLOCKS_PER_SEC << endl;127 128         cout << approximation << endl;129 130     }131     return 0;132 }

 

SOJ 1050. Numbers & Letters