1048 Find Coins (25 point(s))

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本篇博客介绍了一个有趣的算法问题“FindCoins”，即在限定条件下寻找两枚硬币以支付特定金额。通过使用哈希表，博主提供了一种高效的方法来解决这个问题，并分享了实现过程中的关键步骤和潜在的陷阱。

1048 Find Coins (25 point(s))

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=Mand 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

思想：Hash表

题意：在一个数组中寻找恰好两个数（不超过500），让其组成恰好等于M（不超过1000）。并由小到大输出这两个数，如果多组解，则输出第一个数小的解。

基本思路：构建哈希表，记录每一个数的个数cnt[i]。由小到大遍历哈希表，如果cnt[i]>=1&&cnt[M-i]>=1，则得到结果。注意判断两个相同的硬币组成M的情况。

踩坑：case#3段错误（23分）

由于数组大小刚好是500，在判断的时候M-i可能大于500，因此出现段错误。解决方案：1. 数组长度改为1001；2. 判断M-i<=500。

#include<iostream>
using namespace std;
const int MAX = 507;
int cnt[MAX]={0};
int main(void){
	int N,M,a;cin>>N>>M;
	for(int i=0;i<N;i++){
		cin>>a;
		cnt[a]++;
	}
	bool flag=false;
	for(int i=1;i<=M;i++){
		if(i+i==M){
			if(cnt[i]>=2){
				cout<<i<<" "<<i<<endl;
				flag=true;
				break;
			}
		}
		else{
			if(cnt[i]>=1&&M-i<=500&&cnt[M-i]>=1){
				cout<<i<<" "<<M-i<<endl;
				flag=true;
				break;
			}
		}
	}
	if(!flag) cout<<"No Solution"<<endl;
	return 0;
}