1048 Find Coins （25 分）（map以及STL的应用）

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本文探讨了一种特殊的支付场景，即使用两枚硬币精确支付特定金额的问题。通过分析大量硬币面额，利用map和STL进行高效查找，确保能够快速确定是否存在满足条件的两枚硬币。

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1048 Find Coins （25 分）（map以及STL的应用）

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

分析

本题思路较为常规，将数v[i]存在数组v中，同时用map存储v[i]的个数，用于后续查询m-v[i]是否存在，这里采取空间换时间降低时间复杂度,同时在判断m-v[i]是否存在时分两种情况判断，情况1 当v[i]!=m-v[i]时，只要m-v[i]元素个数大于等于1则表明它存在，当v[i]==m-v[i]时，则需要判断m-v[i]元素个数大于等于2时才能表明它存在，因为当m为偶数时，v[i]自身与自身相加也可能等于m，但是此时指存在该数本身，这样就不符合题目要求了~~

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
map<int,int> mp;
vector<int> v;
int main(){
	int n,m;
	cin>>n>>m;
	v.resize(n);
	for(int i=0;i<n;i++){
		cin>>v[i];
		mp[v[i]]++;
	}
	sort(v.begin(),v.end());
	for(auto it : v){
		if((it!=m-it&&mp[m-it]>=1)||(it==m-it&&mp[m-it]>=2)){
			cout<<it<<" "<<m-it<<endl;
			return 0;
		}
	}
	cout<<"No Solution"<<endl;
	return 0;
}