PAT甲级 1048 Find Coins（25 分）hash散列/map-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_29762941/article/details/82467553

本文探讨了FindCoins问题，这是一个关于在大量硬币中寻找两枚特定硬币以支付确切金额的问题。通过使用数组和map两种方法，文章详细介绍了如何高效解决此问题，包括输入输出规格、样例输入输出及代码实现。

1048 Find Coins（25 分）

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：给出n个正整数和一个正整数m，问n个数字中是否存在一对数字a和b(a <= b),使a+b=m成立。如果有多个，输出a最小的那一对。
分析：建立数组arr保存每个数字出现的次数，然后判断输出. 建议用hasp而不是用map。有些题可能会超时。

#include <iostream>
#include <stdio.h>
#include <map>
using namespace std;

int main(){
	int n,m,num;
	scanf("%d %d",&n,&m);
	int arr[100001];
	for(int i=0;i<n;i++){
		scanf("%d",&num);
		arr[num] += 1;
	}
	
	for(int i=1;i<m;i++)
	{
		if(arr[i] >= 1){
			arr[i]--;//先减一，避免两个硬币是同一个值 
			if(arr[m-i] >= 1)
			{
				printf("%d %d",i,m-i);	
				return 0;
			}
			arr[i]++;
		}
	}
	
	printf("No Solution");
	
	return 0;
}

解法二:map

#include <iostream>
#include <stdio.h>
#include <map>
using namespace std;

int main(){
	int n,m,num;
	scanf("%d %d",&n,&m);
	map<int,int> mp;
	for(int i=0;i<n;i++){
		scanf("%d",&num);
		mp[num] += 1;
	}
	
	for(auto it:mp)
	{
		auto other = mp.find(m - it.first);
		if(other != mp.end()){
			if(it.first == other->first && it.second ==1)
				continue;
			if(it.first == other->first && it.second >=2){
				printf("%d %d",it.first,it.first);
				return 0;
			}else{
				printf("%d %d",it.first,other->first);	
				return 0;
			}
		}
	}
	
	printf("No Solution");
	
	return 0;
}