ZCMU 2177 Lucky Numbers (easy)

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http://acm.zcmu.edu.cn/JudgeOnline/problem.php?id=2177

2177: Lucky Numbers (easy)

时间限制: 2 Sec  内存限制: 256 MB
提交: 42  解决: 21
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题目描述

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day Petya came across a positive integer n. Help him to find the least super lucky number which is not less than n.

Input

The only line contains a positive integer n (1≤n≤10⁹). This number doesn't have leading zeroes.

Output

Output the least super lucky number that is more than or equal to n.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

4500

Output

4747

Input

47

Output

47

题目意思：

为你大于等于n的第一个幸运数字是什么

幸运数字：该数字只由4和7组成，且4和7的个数相等

做法：

直接暴力打表（打了大概三分钟，GG）

code：

/*#include <stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include <iostream>
#include<algorithm>*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[]={,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
/*bool f(LL x)
{
    int c1=0,c2=0;
    while(x)
    {
        int a=x%10;
        if(a!=4&&a!=7)
            return false;
        if(a==4)
            c1++;
        if(a==7)
            c2++;
        x/=10;
    }
    if(c1==c2)
        return true;
    else
        return false;
}*/
int main()
{
   /* for(LL i=47;i<10000000000;i++)
    {
        if(f(i))
        {
            cout<<i<<',';
        }
    }
    printf("\n*******\n");*/
    LL n;
    cin>>n;
    for(int i=;;i++)
    {
        if(a[i]>=n)
        {
            cout<<a[i]<<endl;
            break;
        }
    }
    return ;
}