Binary String Constructing CodeForces - 1003B （构造）_re given three integers aa, bb and xx. your task…-优快云博客

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给定三个整数a、b和x，需要构造一个长度为n=a+b的二进制字符串，其中包含a个0，b个1，并且有x个位置i满足si不等于si+1。保证总有一种解决方案。可以通过交替输出0和1来达到x个不同的相邻字符，然后将剩余的0和1连续输出以满足条件。

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You are given three integers $a$ , $b$ and $x$ . Your task is to construct a binary string $s$ of length $n = a + b$ such that there are exactly $a$ zeroes, exactly $b$ ones and exactly $x$ indices $i$ (where $1 \leq i < n$ ) such that $s_{i} \neq s_{i + 1}$ . It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices $i$ such that $1 \leq i < n$ and $s_{i} \neq s_{i + 1}$ ( $i = 1, 2, 3, 4$ ). For the string "111001" there are two such indices $i$ ( $i = 3, 5$ ).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers $a$ , $b$ and $x$ ( $1 \leq a, b \leq 100, 1 \leq x < a + b)$ .

Output

Print only one string $s$ , where $s$ is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

   Examples 
 
     input 
    
      Copy 
    
2 2 1

     output 
    
      Copy 
    
1100

     input 
    
      Copy 
    
3 3 3

     output 
    
      Copy 
    
101100

     input 
    
      Copy 
    
5 3 6

     output 
    
      Copy 
    
01010100

Note

All possible answers for the first example:

1100;
0011.

All possible answers for the second example:

110100;
101100;
110010;
100110;
011001;
001101;
010011;

001011.

题意：给出a，b，x，构造一个01串，有a个0，b个1。这个01串刚好有x个 ai 使得ai！=ai+1。

思路：对于x，我们很容易就可以想到先输出x/2对0和1（n对0和1交错出现可以提供2*n-1个符合题意的ai），然后将剩余的0和剩余的1连续输出（提供1个符合条件的ai，这样就刚好是x个符合条件的ai了，需要注意的是，我们要优先将个数多的放在前面，例如，有10个1,5个0的话，我们先输出x/2个“10”，否则，输出x/2个“01”）。

#include "iostream"
using namespace std;
int main()
{
    int a,b,x,n;
    char u,v;
    cin>>a>>b>>x;
    n=a+b;
    if(a>b) u='0',v='1';
    else {u='1',v='0';swap(a,b);}
    for(int i=0;i<x/2;i++) {cout<<u<<v;a--,b--;}
    if(x%2==0){
        while(b--) cout<<v;
        while(a--) cout<<u;
    }
    else{
        while(a--) cout<<u;
        while(b--) cout<<v;
    }
    return 0;
}