Educational Codeforces Round 81 (Rated for Div. 2) A. Display The Number

原题：You have a large electronic screen which can display up to 998244353998244353 decimal digits. The digits are displayed in the same way as on different electronic alarm clocks: each place for a digit consists of 77 segments which can be turned on and off to compose different digits. The following picture describes how you can display all 1010 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 11, you have to turn on 22 segments of the screen, and if you want to display 88, all 77 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 nn 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 nn segments.

Your program should be able to process tt different test cases.

Input
The first line contains one integer tt (1≤t≤1001≤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 nn (2≤n≤1052≤n≤105) — the maximum number of segments that can be turned on in the corresponding testcase.

It is guaranteed that the sum of nn over all test cases in the input does not exceed 105105.

Output
For each test case, print the greatest integer that can be displayed by turning on no more than nn segments of the screen. Note that the answer may not fit in the standard 3232-bit or 6464-bit integral data type.

Example
inputCopy
2
3
4
outputCopy
7
11
思路：水题。把每一个数字对应几段写出来，重复的留下一个相同段数数字最大的，剩下的里面把能用小的数字组成的数字删掉（如7=2+2+3）最后留下的就是1对应2段，7对应3段，把数判有最多几个2和几个3就好，数字位数越多越大。

#include
#include
#include
#include
#include
#include
#include
#include
#include 
#include
#include
#include
#include 
#include
using namespace std;
int main()
{
    int n,t,a,b,c;
	cin>>t;
	while(t--){
		cin>>n;
		int cnt=0;
		while(n>3){
			n-=2;
			cnt++;
		}
		if(n==3){
			cout<<7;
		}
		else cout<<1;
		for(int i=0;i<cnt;i++){
			cout<<1;
		}
		cout<<endl;
	} 
}