SOJ 1050. Numbers & Letters

题目大意：给出5个整数和4种运算(加法，减法，乘法和除法)，选取其中的部分数作任意运算，使得值等于目标数或者最接近目标数。其中，除法只能整除，即商必须是整数。

解题思路：每次选取任意两个数，进行可行的运算，获得结果并且和目标数作比较。最后得出答案。

代码如下：

  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 }