PAT 甲级 1048 Find Coins

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本文介绍了一个算法问题，旨在从大量硬币中找到两个面值之和等于特定金额的组合。通过预处理和排序，该问题可以在合理的时间内解决，适用于如宇宙购物中心等场景，确保能够使用恰好两枚硬币支付任何账单。

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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=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

经验总结：

这一题....和很多题都很像，比如PAT 乙级 1090 危险品装箱以及 PAT 乙级 1065 单身狗，不过这一题要注意这么一种情况，就是比如第二个样例，14，而money中只有一个7，至少需要两个才够，所以标记数组要记录数量，而不是是否出现了，其他的就没什么难度啦~(*•̀ㅂ•́)و

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1010;
int a[100010];
int flag[maxn]={0};
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<n;++i)
	{
		scanf("%d",&a[i]);
		if(a[i]<=m)
		{
			++flag[a[i]];
		}
	}
	sort(a,a+n);
	int f=0;
	for(int i=0;i<n;++i)
	{
		if(a[i]==m-a[i]&&flag[a[i]]>=2||a[i]!=m-a[i]&&flag[a[i]]&&flag[m-a[i]])
		{
			f=1;
			printf("%d %d\n",a[i],m-a[i]);
			break;
		}
	}
	if(f==0)
		printf("No Solution\n");
	return 0;
}