输出固定个数的相邻两位数不同的字符串

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.

# 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

输出固定个数的相邻两位数不同的字符串

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