CF 1295——A.Display The Number【构造、简单思维】

题目传送门

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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.

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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.

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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.

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input

2
3
4

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output

7
11

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题意

• 每个数字如图由木棒组成，给出一定数量的木棒，求出可以组成的最大数

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题解

• 很显然 $2$ 个木棒即可组成 $1$， $3$ 个木棒组成 $7$，位数越多，数字越大
• 偶数直接全转换成 $1$
• 奇数将第一个 $1$，变成 $7$ 即可

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AC-Code

#include <bits/stdc++.h>
using namespace std;

int main() {
	int T;
	cin >> T;
	while (T--) {
		int n;
		cin >> n;
		int	t = 0;
		if (!(n & 1))
			t = n / 2;
		else
			t = (n - 3) / 2;
		n = n - t * 2;
		if (n)
			printf("7");
		for (int i = 1; i <= t; ++i)
			printf("1");
		puts("");
	}
	return 0;
}

来源：CSDN

作者：nirvana · rebirth

链接：https://blog.csdn.net/Q_1849805767/article/details/104118137