My first solved-problem on Codeforces.↵
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It's really an easy problem.↵
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As we know, if you want to make the a and b as small as possible, they must only contain digits 0 and 1.↵
↵
So, we have the thinking in the following:↵
↵
1. if the digit of x[i] is 2: we try to make a[i] and b[i] both 1.↵
2. if the digit of x[i] is 0: we try to make a[i] and b[i] both 0.↵
3. if the digit of x[i] is 1: either a[i] or b[i] would be 1. We'll make a[i] 1 because it is no difference if we make b[i] 1.↵
↵
However, if we make a[i] 1, the number a must be larger than b as a fact.↵
↵
So, we would not want to make array a still bigger. So after we make a[i] 1, we have different thinking:↵
↵
1. if the digit of x[i] is 2: we make a[i] 0 and b[i] 2.↵
2. if the digit of x[i] is 0: we make a[i] and b[i] both 0.↵
3. if the digit of x[i] is 1: we make a[i] 0 and b[i] 1.↵
↵
That's all. We'll discover that after we make a[i] 1, a does not increase anymore, but b is increasing, however it will never be bigger than a.↵
↵
Thanks for reading, hope that'll help you!
↵
It's really an easy problem.↵
↵
As we know, if you want to make the a and b as small as possible, they must only contain digits 0 and 1.↵
↵
So, we have the thinking in the following:↵
↵
1. if the digit of x[i] is 2: we try to make a[i] and b[i] both 1.↵
2. if the digit of x[i] is 0: we try to make a[i] and b[i] both 0.↵
3. if the digit of x[i] is 1: either a[i] or b[i] would be 1. We'll make a[i] 1 because it is no difference if we make b[i] 1.↵
↵
However, if we make a[i] 1, the number a must be larger than b as a fact.↵
↵
So, we would not want to make array a still bigger. So after we make a[i] 1, we have different thinking:↵
↵
1. if the digit of x[i] is 2: we make a[i] 0 and b[i] 2.↵
2. if the digit of x[i] is 0: we make a[i] and b[i] both 0.↵
3. if the digit of x[i] is 1: we make a[i] 0 and b[i] 1.↵
↵
That's all. We'll discover that after we make a[i] 1, a does not increase anymore, but b is increasing, however it will never be bigger than a.↵
↵
Thanks for reading, hope that'll help you!
