A. Puzzle From the Future

In the 2022 year, Mike found two binary integers a and b of length n (both of them are written only by digits 0 and 1) that can have leading zeroes. In order not to forget them, he wanted to construct integer d in the following way:

he creates an integer c as a result of bitwise summing of a and b without transferring carry, so c may have one or more 2-s. For example, the result of bitwise summing of 0110 and 1101 is 1211 or the sum of 011000 and 011000 is 022000;
after that Mike replaces equal consecutive digits in c by one digit, thus getting d. In the cases above after this operation, 1211 becomes 121 and 022000 becomes 020 (so, d won’t have equal consecutive digits).
Unfortunately, Mike lost integer a before he could calculate d himself. Now, to cheer him up, you want to find any binary integer a of length n such that d will be maximum possible as integer.

Maximum possible as integer means that 102>21, 012<101, 021=21 and so on.

Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.

The first line of each test case contains the integer n (1≤n≤105) — the length of a and b.

The second line of each test case contains binary integer b of length n. The integer b consists only of digits 0 and 1.

It is guaranteed that the total sum of n over all t test cases doesn’t exceed 105.

Output
For each test case output one binary integer a of length n. Note, that a or b may have leading zeroes but must have the same length n.

解析

为了使其最大，首先使最高位最大并保证这个数最长。

#include<bits/stdc++.h>
char b[100100],c[100100],a[100010];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    int n;
    scanf("%d",&n);
    scanf("%s",b);
    if(b[0]=='1')
    {
      c[0]='2';
    }
    else
    {
      c[0]='1';
    }
    for(int i=1;i<n;i++)
    {
      if(b[i]=='0')
      {
        if('1'==c[i-1])
        {
          c[i]='0';
          a[i]='0';
        }
        else
        {
          c[i]='1';
          a[i]='1';
        }
      }
      else
      {
        if('2'==c[i-1])
        {
          c[i]='1';
          a[i]='0';
        }
        else
        {
          c[i]='2';
          a[i]='1';
        }
      }
    }
    a[0]='1';
    for(int i=0;i<n;i++)
    {
      printf("%c",a[i]);
    }
    printf("\n");
  }
}