Binary String Constructing（CodeForces - 1003B）

Description

You are given three integers aa, bb and xx. Your task is to construct a binary string ss of length n=a+bn=a+bsuch that there are exactly aa zeroes, exactly bb ones and exactly xx indices ii (where 1≤i<n1≤i<n) such that si≠si+1si≠si+1. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices ii such that 1≤i<n1≤i<n and si≠si+1si≠si+1 (i=1,2,3,4i=1,2,3,4). For the string "111001" there are two such indices ii (i=3,5i=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 aa, bb and xx (1≤a,b≤100,1≤x<a+b)1≤a,b≤100,1≤x<a+b).

Output

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

Examples

Input

2 2 1

Output

Input

3 3 3

Output

101100

Input

5 3 6

Output

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，构造一个只含0和1的串，要求有a个0，b个1，并且不多不少刚刚好有x个 a [ i ] 使得a [ i ] ! = a [ i + 1 ] 。

思路：

输入o代表0的个数，e代表1的个数，x代表a[i] ！= a[i+1]的个数，（因为o看起来像0，e读起来像1，防止弄乱出错嘛嘻嘻......）注意字符串最后一个是‘\0’，所以可以先给a清零。然后先看0和1哪个个数居多，就先将第一个设为哪个数，然后优先按照01010101或者10101010顺序排列，根据x的数量将不等的先排了，保证后来不再多增加不相等的个数，所以让a[k-1]和a[k]相等然后一定注意最后剩余的0和1还有一次不等。此处还有一个要注意的地方，就是a数组如果设成int容易出错，最好按字符串处理。最后输出时的判断条件也不要设成 i < e+o，因为e+o已经减没了。

代码：

#include<cmath>
#include<deque>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-10

using namespace std;

int main()
{
    int n,o,e,x,k=1;
    scanf("%d%d%d",&o,&e,&x);   //o代表0的个数，e代表1的个数，x代表a[i] ！= a[i+1]的个数
    char a[o+e];
    n=o+e;                     
    memset(a,0,sizeof(a));      //给a清零，注意字符串最后一个是‘\0’
    if(o<=e){
        a[0]='1';
        e--;
    }
    else{                      //看0和1哪个个数居多，就先将第一个设为哪个数
        a[0]='0';
        o--;
    }
    while(x>1)                 //注意最后剩余的0和1还有一次不等，所以此处不能取等
    {
        x--;
        if(a[k-1]=='1'){
            a[k++]='0';        //根据x的数量，优先按照01010101或者10101010顺序排列
            o--;
        }
        else{
            a[k++]='1';
            e--;
        }
    }
    if(a[k-1]=='1'){           //保证后来不再多增加不想等的个数，所以让a[k-1]和a[k]相等
        while(e--)
            a[k++]='1';
        while(o--)
            a[k++]='0';
    }
    else{
        while(o--) 
            a[k++]='0';
        while(e--)
            a[k++]='1';
    }
    for(int i=0;i<=n-1;i++)
        printf("%c",a[i]);
    printf("\n");
    return 0;
}

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