HDU 5676 ztr loves lucky numbers

—亚信科技，巴卡斯（杭州），壹晨仟阳(杭州)，英雄互娱（杭州） 
（包括2016级新生）除了校赛，还有什么途径可以申请加入ACM校队？ 

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 51    Accepted Submission(s): 24

Problem Description

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

 

Input

There are T(1≤n≤105) cases

For each cases:

The only line contains a positive integer n(1≤n≤1018). This number doesn't have leading zeroes.

 

Output

For each cases
Output the answer

 

Sample Input

2
4500
47

 

Sample Output

4747
47

把1到10^20中满足条件的全部枚举出来，然后再二分查找就可以了。
注意20位最小的44444444447777777777在是超long long int 的
只好进行一下特判
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
#include <map>

using namespace std;
#define MAX 600000
typedef __int64 LL;
LL n;
LL a[MAX];
int cnt;
void dfs(LL x,LL y,LL num)
{
    if(x==0&&y==0)
    {
        a[++cnt]=num;
        return;
    }
    if(x>0)
       dfs(x-1,y,num*10+4);
    if(y>0)
       dfs(x,y-1,num*10+7);
}
void fun()
{
    a[1]=47;
    a[2]=74;
    cnt=2;
    for(int i=4;i<=18;i+=2)
    {
        dfs(i/2,i/2,0);
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    fun();
    sort(a+1,a+cnt+1);
    while(t--)
    {
       scanf("%I64d",&n);
       if(n>777777777444444444)
       {
           cout<<"44444444447777777777"<<endl;
           continue;
       }
       LL l=1,r=cnt;
       LL res;
       while(l<=r)
       {
           LL mid=(l+r)/2;
           if(a[mid]>=n)
           {
               res=a[mid];
               r=mid-1;
           }
           else
               l=mid+1;
       }
       printf("%I64d
",res);
    }
    return 0;
}