【PAT】1048. Find Coins (25)

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本文提供了一道 PAT-A 1048 编程题的详细解答，介绍了如何使用两种不同的算法策略来寻找能够支付指定金额的两枚硬币的面值。一种方法利用了哈希表进行快速查找，另一种则是通过排序和双指针技巧实现。

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1048

题目描述：

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：（1）用查询表，即 input[i] 标记数i出现的次数。（2）要考虑M等于某个数的两倍的情况。比如14 = 7 + 7。

#include<iostream>
#include<string.h>
using namespace std;

#define max 100000
int input[max];

int main()
{
	int N,M;
	cin>>N>>M;
	int i,j;
	int temp;
	int flag;
	bool find = false;
	memset(input,0,sizeof(input));
	for(i=0; i<N; i++)
	{
		cin>>temp;
		input[temp] ++;		
	}
	for(i=0; i<=M/2; i++)
	{
		//以下两行是为了判断是否存在两个同样的数和为题目所求，比如14=7+7；
		input[i]--;
		input[M-i]--;
		if(input[i]>=0 && input[M-i]>=0)
		{
			find = true;
			cout<<i<<" "<<M-i<<endl;
			break;
		}
	}

	if( !find )
		cout<<"No Solution"<<endl;
	return 0;
}

另附其他的一些参考代码。

（1）http://blog.163.com/hgfeaon@126/blog/static/95856212013119113055361/

#include <iostream>  

#include <vector>  

#include <algorithm>  

using namespace std;    

int main() {

      int n,m;

      cin>>n>>m;

      vector<int> coins(n);

      while (n--) {

          cin>>coins[n];

      }

      sort(coins.begin(), coins.end());

      vector<int>::iterator iter;

      for(iter = coins.begin(); iter != coins.end(); iter++) {

          int other = m - *iter;

          if (binary_search(iter + 1, coins.end(), other)) {

              cout<<*iter<<" "<<other<<endl;

              break;

          }

      }

      if (iter == coins.end()) {

          cout<<"No Solution"<<endl;

      }

      return 0;

  }

#include <iostream>  

#include <vector>  

#include <algorithm>    

using namespace std;    

int main() {

      int n, m;

      cin>>n>>m;

      vector<int> coins(n);

      while (n--) {

          cin>>coins[n];

      }

      sort(coins.begin(), coins.end());

      int i = 0;

      int j = coins.size() - 1;

      while(i < j) {

          int value = coins[i] + coins[j];

          if (value > m) {

              j--;

          } else if (value < m) {

              i++;

          } else {

              break;

          }

      }

      if (i < j) {

          cout<<coins[i]<<" "<<coins[j]<<endl;

      } else {

          cout<<"No Solution"<<endl;

      }

      return 0;

  }

（2）http://blog.youkuaiyun.com/sunbaigui/article/details/8656978

#include<iostream>
#include<vector>
#define FaceMax 1000

int main()
{
	int n, m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		//input
		std::vector<int> coins;
		coins.assign(FaceMax+1, 0);
		while(n--)
		{
			int tmp;
			scanf("%d",&tmp);
			coins[tmp]++;
		}
		//search
		bool flag = false;
		for(int i = 1; i <= m/2; ++i)
		{
			coins[i]--;
			coins[m-i]--;
			if(coins[i]>=0 && coins[m-i] >= 0)
			{
				printf("%d %d\n",i,m-i);
				flag = true;
				break;
			}
		}
		//no solution
		if(!flag)
			printf("No Solution\n");
	}
}