CodeForces 1295 A Display The Number

See this article on my own blog https://dyingdown.github.io/2020/01/30/CodeForces-1295A-Display-The-Number/

A. Display The Number time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output:standard 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:

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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 $t(1\le t\le 100)$ — 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 $n(2\le n\le 105)$ — 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

input

2
3
4

output

7
11

Analysis

The more number it can lighten, the longer digits it can have. So,

$1$ uses the less segments.

We make $1$ as many as possible. If there are still segments left that can’t make a complete $1$, we break up one $1$ and let it form $7$ with the rest.(the rest could only be 1 or 0 because $1$ uses two segments)

Code

#include<bits/stdc++.h>
 
using namespace std;
 
int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int digits1 = n / 2;
        if(n % 2 == 1) {
            cout << 7;
            for(int i = 1; i < digits1; i ++) {
                cout << 1;
            }
        } else {
            for(int i = 0; i < digits1; i ++) {
                cout << 1;
            }
        }
        cout << endl;
    }
    return 0;
}