输出固定个数的相邻两位数不同的字符串

B. Binary String Constructing

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $1\le i<n$) such that $s_i\ne s_{i+1}$. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices $i$ such that $1\le i<n$ and $s_i\ne s_{i+1}$ ($i=1,2,3,4$). 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 $x$ ($1\le a,b\le 100,1\le x<a+b)$.

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.

# include<bits/stdc++.h>
using namespace std;
int p[2],n,k,i;
int main()
{
    cin>>p[0]>>p[1]>>n;
    k=p[1]>p[0];
    for(i=1; i<n; i++,p[k]--,k^=1)  //k在0和1之间来回变换
        cout<<char(k+'0');
    cout<<string(p[k],k+'0')<<string(p[!k],!k+'0');  
    //重复输出第一个参数个第二个参数的字符
}

#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;

}

第一种方法，因为后面要输出一堆0和1的时候肯定有一个满足条件即连续输出完1（或0）还要连续输出0（或1），此时中间肯定有相邻两位不同，因此前面计算要少一个。

第二种方法，输出2/x对10或01，若x为偶数，则只差一个，如x=4,输出1010后只满足三个，1010后面就应该是连续的0和连续的1。若x为奇数，则差两个,如x=3,输出10后只满足1个，10后面就应该为连续的1和连续的0