题目描述

For some binary string s (i.e. each character si is either ‘0’ or ‘1’), all pairs of consecutive (adjacent) characters were written. In other words, all substrings of length 2 were written. For each pair (substring of length 2), the number of ‘1’ (ones) in it was calculated.

You are given three numbers:

n0 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 0;
n1 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 1;
n2 — the number of such pairs of consecutive characters (substrings) where the number of ones equals 2.
For example, for the string s=”1110011110”, the following substrings would be written: “11”, “11”, “10”, “00”, “01”, “11”, “11”, “11”, “10”. Thus, n0=1, n1=3, n2=5.

Your task is to restore any suitable binary string s from the given values n0,n1,n2. It is guaranteed that at least one of the numbers n0,n1,n2 is greater than 0. Also, it is guaranteed that a solution exists.

Input
The first line contains an integer t (1≤t≤1000) — the number of test cases in the input. Then test cases follow.

Each test case consists of one line which contains three integers n0,n1,n2 (0≤n0,n1,n2≤100; n0+n1+n2>0). It is guaranteed that the answer for given n0,n1,n2 exists.

Output
Print t lines. Each of the lines should contain a binary string corresponding to a test case. If there are several possible solutions, print any of them.

样例

7
1 3 5
1 1 1
3 9 3
0 1 0
3 1 2
0 0 3
2 0 0

1110011110
0011
0110001100101011
10
0000111
1111
000

这道题的思路就是构造一个符合条件的01串 有个点 当n1等于0时候一定会有 n0 和 n2 其中一个等于0

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n0,n1,n2;
        cin>>n0>>n1>>n2;
        if(n1==0)
        {
            if(n0)
            {
                for(int i=1;i<=n0+1;i++)
                cout<<0;
                cout<<endl;
            }
            else{
                for(int i=1;i<=n2+1;i++)
                cout<<1;
                cout<<endl;
            }
        }
        else{
            for(int i=1;i<=n0+1;i++)
            cout<<0;
            n1--;
            for(int i=1;i<=n2+1;i++)
            cout<<1;
            for(int i=1;i<=n1;i++)
            {
                if(i&1) cout<<0;
                else cout<<1;
            }
            cout<<endl;
        }
    }
    return 0;
}