**You have a large electronic screen which can display up to 998244353 decimal digits. The digits are displayed in the same way as on different electronic alarm clocks: each place for a digit consists of 7 segments which can be turned on and off to compose different digits. The following picture describes how you can display all 10 decimal digits:
As you can see, different digits may require different number of segments to be turned on. For example, if you want to display 1, you have to turn on 2 segments of the screen, and if you want to display 8, all 7 segments of some place to display a digit should be turned on.
You want to display a really large integer on the screen. Unfortunately, the screen is bugged: no more than n segments can be turned on simultaneously. So now you wonder what is the greatest integer that can be displayed by turning on no more than n segments.
Your program should be able to process t different test cases.
Input
The first line contains one integer t (1≤t≤100) — the number of test cases in the input.
Then the test cases follow, each of them is represented by a separate line containing one integer n (2≤n≤105) — the maximum number of segments that can be turned on in the corresponding testcase.
It is guaranteed that the sum of n over all test cases in the input does not exceed 105.
Output
For each test case, print the greatest integer that can be displayed by turning on no more than n segments of the screen. Note that the answer may not fit in the standard 32-bit or 64-bit integral data type.
Example
inputCopy
2
3
4
outputCopy
7
11
题意:
0~9数字由不同数目的火柴棒组成，现在题目给出火柴棒数量，让你用这些火柴棒组成一个最大数。
解析:
位数越多，数就越大
我们找到1，1是最少木棒组成的。但这还不够加入现在给你7个木棒全部用1的话，答案是111，是错误的。正确答案是711.所以我们还需要判断n是不是奇数如果是输出一个7，剩下的都输出1即可。
#include<bits/stdc++.h>
using namespace std;
map<int,int> v;
int t;
int main()
{
cin>>t;
int x;
while(t--)
{
cin>>x;
if(x&1)
{
cout<<7;
x-=3;
}
while(x)
{
cout<<1;
x-=2;
}
cout<<endl;
}
}
