zoukankan      html  css  js  c++  java
  • Flea travel


    Description

    Can you imagine our life if we removed all zeros from it? For sure we will have many problems.

    In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation a + b = c, where a and b are positive integers, and c is the sum of a and b. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?

    For example if the equation is 101 + 102 = 203, if we removed all zeros it will be 11 + 12 = 23 which is still a correct equation.

    But if the equation is 105 + 106 = 211, if we removed all zeros it will be 15 + 16 = 211 which is not a correct equation.

    Input

    The input will consist of two lines, the first line will contain the integer a, and the second line will contain the integer b which are in the equation as described above (1 ≤ a, b ≤ 109). There won't be any leading zeros in both. The value of c should be calculated as c = a + b.

    Output

    The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.

    Sample Input

    Input

    101102

    Output

    YES

    Input

    105106

    Output

    NO

    题意 输入两个数a和b 求这ab之和c   把abc中的0去掉得到新的abc,新的ab之和等于新的c则输出"YES"否则输出"NO"

    思路  自定义一个函数使得输入一个数能够得到去0之后的数

                int shi(int x)
    {
     int shu=0,y=x,ci=1;
     while(y)
     {
      int yu=y%10;       <不断取该数的最后一位>
      if(yu)                       <判断是否为0>
      shu+=ci*yu,ci*=10;      
      y=y/10;               <最后一位已经判断过,将其去掉>
     }
      return shu;
    }

    #include<cstdio>
    int shi(int x)
    {
    	int shu=0,y=x,ci=1;
    	while(y)
    	{
    		int yu=y%10;
    		if(yu)
    		shu+=ci*yu,ci*=10;
    		y=y/10;
    	}
    		return shu;
    }
    	int main()
    	{
    			int a,b,c;
    	scanf("%d%d",&a,&b);
    	c=a+b;
    	if(shi(c)==shi(a)+shi(b))
    	printf("YES
    ");
    	else
    	printf("NO
    "); 
    return 0;
    
    	}
    	


     

  • 相关阅读:
    【06月18日】A股滚动市净率PB历史新低排名
    沪深300指数的跟踪基金最近1年收益排名
    主要股东近3年净买入排名
    北上资金近1周流入排行榜
    【06月12日】指数估值排名
    最近一月研报推荐次数最多的最热股票
    【06月10日】A股ROE最高排名
    JDK源码阅读-------自学笔记(九)(常用类型Integer初探)
    JDK源码阅读-------自学笔记(八)(数组演示冒泡排序和二分查找)
    JDK源码阅读-------自学笔记(七)(二维数组的浅析)
  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027238.html
Copyright © 2011-2022 走看看