【PAT】1048. Find Coins (25)

題目連結：http://pat.zju.edu.cn/contests/pat-a-practise/1048

題目描述：

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）用查詢表，即 input[i]  标記數i出現的次數。（2）要考慮M等于某個數的兩倍的情況。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++)
	{
		//以下兩行是為了判斷是否存在兩個同樣的數和為題目所求，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;
}
           

另附其他的一些參考代碼。

（1）http://blog.163.com/[email protected]/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.csdn.net/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");
	}
}