PAT 1048 Find Coins

原创于 2021-11-24 22:05:47 发布 · 155 阅读

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这是一个关于编程算法的问题，输入包含硬币总数N和目标金额M，需要找出两个硬币面值V1和V2，使得V1+V2=M，并且V1尽可能小。给定的AC代码实现了该功能，通过遍历硬币并使用哈希映射来检查解决方案。当没有找到合适的硬币组合时，输出NoSolution。

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

结尾无空行

Sample Output 1:

4 11

结尾无空行

Sample Input 2:

7 14
1 8 7 2 4 11 15

结尾无空行

Sample Output 2:

No Solution

题目分析

芜湖~~

AC代码

#include <iostream>
#include <vector>
#include <map>
using namespace std;
vector<int> coins;
map<int, int> book;
int main() {
    int N, target;
    cin >> N >> target;
    int temp;
    int flag = 0;
    int ans = target+2;
    for (int i = 0; i < N; i++) {
        cin >> temp;
        coins.push_back(temp);
    }
    for (int i = 0; i < N; i++) {
        if (book.find(coins[i]) != book.end()) {
            if (target - coins[i] < ans) {
                ans = min(target - coins[i],coins[i]);
                flag = 1;
            }
        }
        book[target - coins[i]] = i;
    }
    if (flag == 0) {
        cout << "No Solution";
        return 0;
    }
    cout << ans << ' ' << target - ans;
    return 0;
}