PAT甲级——A1048 Find Coins-白红宇

PAT甲级——A1048 Find Coins

阅读量：4540 次

发布时间：2019-06-08

本文共 2635 字，大约阅读时间需要 8 分钟。

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as $1 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.$

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers:

Output Specification:

For each test case, print in one line the two face values

Sample Input 1:

8 151 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 141 8 7 2 4 11 15

Sample Output 2:

No Solution

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 using namespace std; 6 int N, M; 7 //放法一，使用窗口函数 8 int main() 9 {10     cin >> N >> M;11     vector
         
          nums(N);12     vector
          
           
            >res;13 for (int i = 0; i < N; ++i)14 cin >> nums[i];15 sort(nums.begin(), nums.end(), [](int a, int b) { return a < b; });16 int l = 0, r = N - 1, m;17 while (l < r)//找到中间的数刚好与M/2最接近18 {19 m = (l + r) / 2;20 if (nums[m] <= M / 2 && nums[m + 1] > M / 2)21 break;22 if (nums[m] > M / 2)23 r = m - 1;24 else25 l = m + 1;26 }27 r = m + 1;28 l = m;29 while(l>=0 && r
            
              a, pair
             
               b) { return a.first < b.first; });42 if (res.size() == 0)43 cout << "No Solution" << endl;44 else45 cout << res[0].first << " " << res[0].second << endl;46 return 0;47 }48 49 //方法二，使用数找数原理50 int main()51 {52 cin >> N >> M;53 int nums[1001], t;//根据题目要新建的数组大小54 memset(nums, 0, sizeof(int) * 1001);55 for (int i = 0; i < N; ++i)56 {57 cin >> t;58 nums[t]++;//统计个数59 }60 for (int i = 0; i < 1001; ++i)61 {62 if (nums[i] > 0)//个数大于063 {64 nums[i]--;//使用掉一个数字65 if (M > i && nums[M - i] > 0)66 {67 cout << i << " " << M - i << endl;//由最小值开始遍历，故为最优答案68 return 0;69 }70 nums[i]++;//没有使用，还回去71 }72 }73 cout << "No Solution" << endl;74 return 0;75 }

转载于:https://www.cnblogs.com/zzw1024/p/11275777.html

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