1048 Find Coins (25 分)

本文探讨了一个有趣的算法问题:如何从大量的硬币中找到两枚硬币,其总价值恰好等于给定的金额。通过分析输入数据,使用有效的数据结构和算法,如二分查找和哈希表,来解决这一挑战。

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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

Sample Output 1:

4 11

Sample Input 2:

7 14
1 8 7 2 4 11 15

Sample Output 2:

No Solution

 自己的代码:

#include <iostream>
#include <vector>
#include <algorithm>	

using namespace std;


int n, payment;
vector<int > v;



bool find(int target)
{
	int left = 0, right = n - 1;
	
	while (left<=right)
	{
		int mid = (left + right) / 2;
		if (v[mid]==target)
		{
			return true;
		}
		else if(v[mid]<target)
		{
			left = mid + 1;
		}
		else
		{
			right = mid-1;
		}
	}
	return false;
}
int main()
{
	//freopen("in.txt", "r", stdin);
	
	int cnt[100001] = { 0 }; //记录那些同一的面值的个数
	cin >> n >> payment;
	for (int i = 0; i < n; i++)
	{
		int temp;
		cin >> temp;
		v.push_back(temp);
		cnt[temp]++;
	}
	sort(v.begin(), v.end());
	for (int i = 0; i < n; i++)
	{
		int a = v[i];//在这里为了解决payment=4,有两个面值为2的硬币这种情况,因为我用的vector嘛,还试着v.erase(it),it是迭代器指向v[i]。 但是,c++primer说vector容量一发生变化,迭代器就失效。p99见c++ primer。
		cnt[a]--;
		
		int target = payment - a;
		if (a==target&&cnt[a]==0)
		{
			continue;
		}
		if (find(target)) //二分查找想着时间复杂度会低,但是柳的代码好像不用查。
		{
			printf("%d %d", a, target);
			return 0;
		}
		v[i] = a;
	}
	cout << "No Solution" << endl;
	//fclose(stdin);
	return 0;
}

 

 柳婼的绝妙代码:

#include <iostream>
#include <vector>
#include <algorithm>	

using namespace std;


int a[1001];//柳婼开1000可能因为(而且我听人说)有个测试点数据量是1000


//bool find(int target)
//{
//	int left = 0, right = n - 1;
//	
//	while (left<=right)
//	{
//		int mid = (left + right) / 2;
//		if (v[mid]==target)
//		{
//			return true;
//		}
//		else if(v[mid]<target)
//		{
//			left = mid + 1;
//		}
//		else
//		{
//			right = mid-1;
//		}
//	}
//	return false;
//}

int main()
{
	//freopen("in.txt", "r", stdin);
	int n, payment;
	cin >> n >> payment;
	for (int i = 0; i < n; i++)
	{
		int temp;
		cin >> temp;
		a[temp]++;
	}
	for (int i = 0; i < 1001; i++)  //巧妙地用下标代表了 那些硬币的值  a[payment-i]就是看payment-i是不是在给的那些硬币里
	{
		if (a[i])
		{
			a[i]--;//解决 payment=4,然后给的硬币有两个面值为2的硬币这种情况  
			if (payment>i&&a[payment-i])
			{
				cout << i << " " << payment - i;
				return 0;
			}
			a[i]++;
		}
	}
	
	cout << "No Solution" << endl;
	//fclose(stdin);
	return 0;
}

 

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