2177-Lucky Numbers (easy) ZCMU

最新推荐文章于 2022-04-08 22:38:03 发布

原创最新推荐文章于 2022-04-08 22:38:03 发布 · 427 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#全排列

练习专栏收录该内容

160 篇文章

订阅专栏

本文介绍了一种特殊类型的数字——超级幸运数，这类数字仅由4和7组成且两者的数量相等。文章详细阐述了如何找到一个不小于给定整数n的最小超级幸运数的方法，并提供了具体的实现代码。

Description：

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

Output

Input

Output

题意：超级幸运数字是只由数字4和7构成并且4和7的个数相同。题目要求输入一个数n，你要找出一个不小n的超级幸运数

思路：分两种情况 1.当给所数n的位数为奇数时，那么只需输出位数为n的位数+1的最小超级幸运数

2.当给所数n的位数为偶数时，用相同位数的最大超幸运数和它比较，若仍小于n，则输出位数为n的位数+2的最小超级幸运数。若大于n，则将相同位数的最小超级幸运数通过全排列找出等于n或大于n的超级幸运数输出

代码：


#include<bits/stdc++.h>
using namespace std;
void print(int len)
{
    int i;
    for(i=1;i<=len;i++)
        printf("4");
    for(i=1;i<=len;i++)
        printf("7");
}
int main()
{
    char s[15],n[15];
    int i,len,r;
    while(~scanf("%s",s))
    {
        len=strlen(s);
        if(len%2==1)
        {
           print((len+1)/2);
           continue;
        }
        for(i=0;i<len/2;i++)
            n[i]='7';
        for(i=len/2;i<len;i++)
            n[i]='4';
        r=strcmp(s,n);
        if(r==1)
            print((len+2)/2);
        else{
            for(i=0;i<len/2;i++)
				n[i]='4';
			 for(i=len/2;i<len;i++)
				n[i]='7';
			do{
			r=strcmp(s,n);
			if(r==0||r==-1)
            {
               for(i=0;i<len;i++)
                  printf("%c",n[i]);
               break;
             }
               }while(next_permutation(n,n+len));
			}
      printf("\n");
    }
    return 0;
}