time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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

水题
奇数的话先出7再出1，偶数一直出1。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN = 1e6 + 5;


int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--){
		int n;
		int flag=0;
		cin>>n;
		if(n%2){
			flag=1;
		}
		while(n){
			if(flag){
				cout<<7;
				n-=3;
				flag=0;
			}
			if(n>=2&&n!=3) {
				cout<<1;
				n-=2;
			}
		}
		cout<<endl;
	}
	return 0;
}