题目传送门
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.
input
2
3
4
output
7
11
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
题解
- 很显然 2 2 2 个木棒即可组成 1 1 1， 3 3 3 个木棒组成 7 7 7，位数越多，数字越大
- 偶数直接全转换成 1 1 1
- 奇数将第一个 1 1 1，变成 7 7 7 即可
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;elset = (n - 3) / 2;n = n - t * 2;if (n)printf("7");for (int i = 1; i <= t; ++i)printf("1");puts("");}return 0;
}