A1048 Find Coins (25 分)

First, the technical summary

1. First, a little at first glance does not understand the subject, title roughly means Xiaoming there are many coins of different denominations, but now he wants to get a new business here face value,
   and the business has a provision, a new coin must you two coins add up to the nominal value to change, that there is a first problem arises, that is, combinations of two will produce a variety of new coins
   which here takes the minimum, then the output of the small front face value. The second problem is the same two coins may occur.
2. So, I started using an array of type bool, int whether or not to adopt the type of look the same coin denomination may appear twice. So there is a test point was not the first time through the submission.
   After correcting the problem is solved.

Second, the reference code

#include<iostream>
using namespace std;
bool hashTable[1000010] = {0};
int main(){
    int N,M;
    cin >> N >> M;
    int num;//用于记录每次输入的硬币面值 
    for(int i = 0; i < N; i++){
        cin >> num;
        hashTable[num] = true; 
    }
    int flag = 0;//用于标记是否有解决方案 
    for(int i = 1; i < M; i++){
        if(hashTable[i] == true){
            if(hashTable[M-i] == true && M-i != i){
                cout << i << " " << M-i;
                flag = 1;
                break;
            }
        }
    }
    if(flag == 0) cout << "No Solution";
    return 0;
}