Life Without Zeros

Life Without Zeros

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.

题意：a+b=c,去除a,b,c中的0，若等式还成立，输出“YES”，否则输出“NO”。

思路：定义一个函数，不断取余，去除0后再相加，最后判断等式是否成立。

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

int change(int m){
int j=0;
int s=0;
int k[25];
while (m){
k[j++]=m%10;
m/=10;
}
for (int i=j-1;i>=0;i--){
if (k[i])
s=s*10+k[i];
}
return s;
}

int main(){
int a,b,c;
int x,y;
while (scanf("%d%d",&a,&b)!=EOF){
c=a+b;
x=change(c);
y=change(a)+change(b);
if (x==y)
printf("YES\n");
else
printf ("NO\n");
}

return 0;
}