点击链接PAT甲级-AC全解汇总

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 10
​5
​​ 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 (≤10
​5
​​ , the total number of coins) and M (≤10
​3
​​ , 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 V_​1​​ and V_​2​​ (separated by a space) such that V_​1​​ +V_​2​​ =M and V_​1​​ ≤V_​2​​ . If such a solution is not unique, output the one with the smallest V_​1​​ . If there is no solution, output No Solution instead.

Sample Input 1:

8 15
1 2 8 7 2 4 11 15
1
2

Sample Output 1:

4 11
1

Sample Input 2:

7 14
1 8 7 2 4 11 15
1
2

Sample Output 2:

No Solution
1

题意：
输入一堆数，找出两个数的和 正好等于输入的一个目标值

我的思路：
把输入的数放到set里，从小到达找目标值与他的差存在不存在。

注意： 可能出现某个值的两倍是目标值，而这个数指输入了一次的情况

我的代码：

#include
using namespace std;

int main(){
    int N,M;
    cin>>N>>M;
    set<int>coins;
    int times[N+1]={0};
    for(int i=0;i<N;i++)
    {
        int t;
        cin>>t;
        coins.insert(t);
        times[t]++;
    }
    for(auto it:coins)
    {
        if(it==M/2&&times[it]<2)continue;//如果是目标值的一半，并且只出现了一次
        if(coins.find(M-it)!=coins.end())
        {
            cout<<it<<" "<<M-it<<endl;
            return 0;
        }
    }
    cout<<"No Solution"<<endl;
    return 0;
}

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