1048 Find Coins

这道题目描述了一位角色需要在购物时使用硬币支付,且每笔交易必须恰好用两枚硬币完成。给定硬币面额和总价,我们需要找出是否存在这样的硬币组合。首先对硬币进行排序,然后遍历硬币,检查是否存在另一枚硬币与当前硬币相加等于总价。如果找到解,输出硬币面额;若没有解,则输出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

Sample Output 1:

4 11

Sample Input 2:

7 14
1 8 7 2 4 11 15

Sample Output 2:

No Solution

题目大意

给你N个张钞票和一个物品的价格sum

求是否存在任意两张钞票加在一起的价格等于该物品价格

如果解不唯一

优先输出拥有面子最小的钞票组合


思路

将票票从小到大排好序,并用apr[price]记录每个面值的出现次数

如果存在 price 与 sum-price都存在,输出即可

注意俩个票票面值相等的情况~


C/C++ 

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int apr[2022] = {0};
    int N,sum,nums[100001];
    cin >> N >> sum;
    for(int z=0;z<N;z++){
        cin >> nums[z];
        apr[nums[z]]++;
    }
    sort(nums,nums+N);
    for(int z=0;z<N;z++){
        if(nums[z]>=sum) break;
        if(apr[sum-nums[z]]){
            if(sum-nums[z]==nums[z] && apr[nums[z]]<2) continue;
            cout << nums[z] << " " << sum-nums[z];
            return 0;
        }
    }
    puts("No Solution");
    return 0;
}

 


### 关于分割硬币的公平分配算法 在计算机科学和数学领域,分割硬币的问题通常可以被建模为一种优化问题或动态规划问题。目标通常是找到一种方式来最小化两个集合之间的差异或者最大化某种公平性标准。 #### 动态规划解决方案 对于分割硬币使其尽可能均匀分布的情况,可以采用动态规划的方法解决此问题。假设我们有一组硬币 `coins` 和它们的价值分别为 `[c1, c2, ..., cn]`,我们需要将其分成两部分使得这两部分价值之差最小[^2]。 以下是基于动态规划的一个实现方案: ```python def min_difference_partition(coins): total_sum = sum(coins) n = len(coins) dp = [[False]*(total_sum//2 + 1) for _ in range(n+1)] # Initialize DP table for i in range(n+1): dp[i][0] = True for i in range(1, n+1): for j in range(1, total_sum//2 + 1): if coins[i-1] <= j: dp[i][j] = dp[i-1][j] or dp[i-1][j-coins[i-1]] else: dp[i][j] = dp[i-1][j] # Find the largest value that can be achieved less than half of total sum for j in range(total_sum//2, -1, -1): if dp[n][j]: return abs((total_sum - j) - j) ``` 该函数通过构建一个二维布尔数组 `dp` 来记录子集总和的可能性,并最终返回能够达到的最大接近一半总和的值,从而计算出两者间的最小差距[^3]。 #### 贪婪算法近似解法 如果追求更高效的解决方案而允许一定的误差范围,则可以考虑贪婪策略。这种方法并不总是能找到最优解,但在某些情况下表现良好。基本思路是从最大面额开始依次选取直到无法再选为止[^4]。 ```python def greedy_divide_coins(coins): coins.sort(reverse=True) group_a = [] group_b = [] for coin in coins: if sum(group_a) < sum(group_b): group_a.append(coin) else: group_b.append(coin) return (group_a, group_b), abs(sum(group_a)-sum(group_b)) ``` 尽管如此,在实际应用中需注意验证其适用性和局限性。
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