PAT甲级 1048. Find Coins (25)

题目：

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 10⁵ 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: N (<=10⁵, the total number of coins) and M(<=10³, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V₁ and V₂ (separated by a space) such that V₁ + V₂ = M and V₁ <= V₂. If such a solution is not unique, output the one with the smallest V₁. If there is no solution, output "No Solution" instead.

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

思路:

这道题不要从根据所有的硬币先排序再查找的角度出发，要从根据硬币的面值进行计数，从面值上进行判断选择。

最大面值是500，但要注意判断v1存在的同时，判断M-v1是否存在需要先判断其是否满足M-v1<=500的条件，否则会出现数组越界。

代码：

#include<iostream>using namespace std;int coin[501] = { 0 };int main(){int N, M;cin >> N >> M;int i,k;for (i = 0; i < N; ++i){cin >> k;++coin[k];}bool flag = 1;i = 1;while (flag && i<=500 && i<=M/2){if (coin[i] != 0)  //有硬币{if (2 * i != M){if ( (M-i<=500) && coin[M - i] != 0){flag = 0;break;}}else{if (coin[i] >= 2){flag = 0;break;}}}++i;}if (!flag){cout << i << " " << M - i << endl;}else{cout << "No Solution" << endl;}system("pause");return 0;}