zoukankan      html  css  js  c++  java
  • PAT-A 1048. Find Coins

    1048. Find Coins

    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
    

    程序代码:

    #include<stdio.h>
    #define MAX 100001
    int coins[MAX]={0};
    int main()
    {
    	int N,M,value;
    	scanf("%d%d",&N,&M);
    	int i=0;
    	for(i=0;i<N;i++)
    	{
    		scanf("%d",&value);
    		coins[value]++;
    	}
    	int tmp;
    	for(i=1;i<=M/2;i++)
    	{
    		if(coins[i]>0)
    		{
    			coins[i]--;
    			tmp = M-i;
    			if(coins[tmp]>0)
    			{
    				printf("%d %d",i,tmp);
    				return 0;
    			}	
    		}
    	}
    	printf("No Solution");
    	return 0;
    }
    
    
  • 相关阅读:
    定时器应用-最终版
    定时器应用-点击按钮,div向右移动
    通过js读取元素的样式
    延时调用
    定时器应用-切换图片的练习
    BOM对象属性定时器的调用
    BOM浏览器对象模型
    键盘移动
    Python-字符串方法
    Python实现注册和登录
  • 原文地址:https://www.cnblogs.com/zhengkang/p/5734402.html
Copyright © 2011-2022 走看看