1003B - Binary String Constructing

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.

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：输入a,b,x，要构造出一个一个01串，有a个0，b个1，01或者10，有x个。

题解：构造从不擅长构造，所以看别人思路，第二天做，不是比赛ac的...

c++：

#include<bits/stdc++.h>
using namespace std;
int a,b,x,c;
int main()
{
    cin>>a>>b>>x;
    if(a>=b)c=0;
    else c=1;
    while(x>1)
    {
        cout<<c;
        if(c) b--;
        else a--;
        c=1-c,x--;
    }
    if (c)
    {
        for(int i=0; i<b; i++)
            cout<<1;
        for(int j=0; j<a; j++)
            cout<<0;
    }
    else
    {
        for(int i=0; i<a; i++)
            cout<<0;
        for(int j=0; j<b; j++)
            cout<<1;
    }
    return 0;
}

python：

a,b,x=map(int,input().split())
s=""
if a>=b:c=0
else:c=1
while x>1:
    s+=str(c)
    if c==0:a-=1
    else:b-=1
    c=1-c;x-=1
if c==0:s+=a*"0"+b*"1"
else:s+=b*"1"+a*"0"

1003B - Binary String Constructing

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