【PAT甲级 - C++题解】1048 Find Coins

1048 Find Coins

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 15
1 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 14
1 8 7 2 4 11 15

Sample Output 2:

No Solution

题意

给定若干硬币以及要支付的金额 m ，要求从这些硬币中找到恰好两个硬币 v1 和 v2 金额之和刚好等于要支付的金额 m 即 v1 + v2 = m 。

如果结果不唯一，则从所有结果中输出 v1 最小的那一对结果。

思路

我们可以从前往后遍历每一个硬币 x ，每次遍历都去判断支付金额 m 与 x 的差值即 m-x 是否能在之前遍历的硬币中找到对应的金额。如果能找到，则判断结果是否唯一，如果不唯一则要保留 v1 最小的那个结果

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 100010, INF = 0x3f3f3f3f;
int n, m;
unordered_set<int> num;
int main()
{
    cin >> n >> m;
    //找到最佳答案
    int v1 = INF, v2 = INF;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        int y = m - x;
        if (num.count(y))
        {
            if (x > y) swap(x, y);
            if (v1 > x)    v1 = x, v2 = y;
        }
        num.insert(x);
    }
    //输出答案
    if (v1 == INF) puts("No Solution");
    else cout << v1 << " " << v2 << endl;
    return 0;
}