题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5676

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 594    Accepted Submission(s): 257

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

Source

BestCoder Round #82 (div.2)

题目大意：

ztr喜欢幸运数字，他对于幸运数字有两个要求
1：十进制表示法下只包含4、7
2：十进制表示法下4和7的数量相等
比如47,474477就是
而4,744,467则不是

现在ztr想知道最小的但不小于n的幸运数字是多少

解题思路：暴力打出所有幸运数~~~~然后二分查找即可。

详见代码。

#include <iostream>
#include <cstdio>

using namespace std;

#define ll long long

ll a[100000];
int k=0;

void dfs(ll ans,int num4,int num7)
{
    if (num4==0&&num7==0)
    {
        a[k++]=ans;
        return;
    }
    if (num4==0)
    {
        dfs(ans*10+7,num4,num7-1);
    }
    else if (num7==0)
    {
        dfs(ans*10+4,num4-1,num7);
    }
    else
    {
        dfs(ans*10+4,num4-1,num7);
        dfs(ans*10+7,num4,num7-1);
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    for (int i=2;i<=18;i+=2)
        dfs(0,i/2,i/2);
    while (t--)
    {
        ll n;
        scanf("%lld",&n);
        if(n>777777777444444444LL)
        {
            printf("44444444447777777777\n");
            continue;
        }
        int l=0,r=k;
        while (r>l)
        {
            int mid=(l+r)/2;
           // cout<<mid<<endl;
            if (a[mid]<n)
            {
                l=mid+1;
            }
            else
            {
                r=mid;
            }
        }
        printf("%lld\n",a[r]);
    }
    return 0;
}