【水模拟】#67 A. Life Without Zeros

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#Life Without Zeros #Codeforces #Beta Round #67

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本文探讨了在数学等式中去除所有零之后，等式是否仍然成立的问题。通过实例分析，展示了如何判断去掉零后等式是否依然正确。

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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 test(s)

input
101
102

output
YES

input
105
106

output
NO

这道题呀，题意是说给你两个数字（然后你自己算出他们的和），如果把加数以及他们的和里面的零都去掉，这个等式是否还成立

那么就直接把去掉零之后的数字都求出来，计算一下看是不是成立就好啦~

咱们模拟一下就好~ 需要注意一下这个弃掉0的函数怎么写就好^_^

int del0(int x)
{
	int ret=0,tmp=x,time=1; //tmp是这个数字，ret是用来存储去掉0的返回值 
	while(tmp)
	{
		int digit=tmp%10;	//获得当前最低位的数字 
		if(digit) ret+=(time)*digit,time*=10;	//如果不是0就加进去，是0的话无视掉 
		tmp/=10;
	}
	return ret;
}

Code：

#include <cstdio>
#include <string>
#include <cstring> 
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

int del0(int x)
{
	int ret=0,tmp=x,time=1;
	while(tmp)
	{
		int digit=tmp%10;
		if(digit) ret+=(time)*digit,time*=10;
		tmp/=10;
	}
	return ret;
}

int main()
{
	int a,b;
	cin>>a>>b;
	int c=a+b;
	//cout<< del0(c)<<": "<<del0(a)<<": "<<del0(b)<<endl;
	if(del0(c)==del0(a)+del0(b))cout<<"YES"<<endl;
	else cout<<"NO"<<endl;
	return 0;
}