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;
}