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

You have a large electronic screen which can display up to

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

As you can see, different digits may require different number of segments to be turned on. For example, if you want to display

You want to display a really large integer on the screen. Unfortunately, the screen is bugged: no more than

Your program should be able to process

Input

The first line contains one integer

Then the test cases follow, each of them is represented by a separate line containing one integer

It is guaranteed that the sum of

Output

For each test case, print the greatest integer that can be displayed by turning on no more than

菜狗我的思路

写了一大堆没用的东西

int a[10] = {0};
const int CNM = 998244353;
void init()
{
 
    a[5] = 9;
    a[3] = 7;
    a[2] = 1;
}
int main()
{
    int t;
    cin >> t;
    init();
    while (t--)
    {
        int n;
        cin >> n;
        if (n<=CNM*2)
        {
            if (n<4)
                printf("%d",a[n]);
            else{
                if (n%2 == 0)
                    for (int i = 1; i <= n/2; i++)
                        printf("1");
                else
                {
                    printf("7");
                    for (int i = 2; i <= n/2; i++)
                        printf("1");
                }
            }
            printf("
");
        }
        else{
            int y = n - CNM*2;
            for (int i = 1; i<=CNM ; i++)
            {
                bool f= false;
                for (int j = 5; j>= 3; j--)
                {
                    if (a[j] == 0) continue;
                    //printf("y = %d
", y);
                    if (y-j + 2>=0)
                    {
                        y-=(j-2);
                        printf("%d", a[j]);
                        f = true;
                        break;
                    }
                }
                if (f) continue;
                printf("1");
            }
            printf("
");
        }
    }
}

大佬思路：其实就是先摆1，因为位数多的肯定大然后多出来的灯就让原来的最前面的那个1变为7就可以了。

#include<iostream>
#include<string>
#include <cstdlib>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include <iomanip>
#include<algorithm>
// #pragma comment(linker, "/STACK:1024000000,1024000000")
// #define pi acos(-1)
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x7f7f7f7f //2139062143
#define INF1 0x3f3f3f3f //1061109567
#define INF2 2147483647
#define llINF 9223372036854775807
#define pi 3.141592653589793//23846264338327950254
#define pb push_back
#define ll long long
#define debug cout << "debug
";
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
#define CNM ios::sync_with_stdio(false);cin.tie(NULL);
#define scai(x) scanf("%d", &x)
#define sca2i(x, y) scanf("%d %d", &x, &y)
#define scaf(x) scanf("%lf", &x)
#define sca2f(x, y) scanf("%lf %lf", &x, &y)
#define For(m,n) for (int i = m;  i < n; i++)

inline int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')w = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
}
#define local
#ifdef local
#endif

#define MAX 10233
#define LCH(i) ((i) << 1)
#define RCH(i) ((i) << 1 | 1)

int main()
{
    CNM;
    int t; cin >> t;
    while (t--)
    {
        int n; cin >> n;
        int cnt = n / 2;
        string s = "";
        for (int i = 0; i < cnt; i++)
        {
            s += "1";
        }

        if (n & 1)
        {
            s = "7" + s;
            s.pop_back();
        }
        cout << s << endl;
    }
}

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

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