hdoj 5676 ztr loves lucky numbers 【二分】

题目链接：hdoj 5676 ztr loves lucky numbers

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 334 Accepted Submission(s): 136

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

题意：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

思路：一开始分块预处理，结果WA了。后来没有分块就过了，一脸懵懂。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 5*1e6 +10;
const int INF = 0x3f3f3f3f;
__int64 ans[500000];
int num;
void DFS(int pos, int cnt1, int cnt2, int len, __int64 sum) {
    if(pos == len) {
        if(cnt1 < cnt2) {
            sum = sum * 10 + 4;
        }
        else {
            sum = sum * 10 + 7;
        }
        ans[num++] = sum;
        return ;
    }
    if(cnt1 + 1 <= len / 2) DFS(pos+1, cnt1+1, cnt2, len, sum * 10 + 4);
    if(cnt2 + 1 <= len / 2) DFS(pos+1, cnt1, cnt2+1, len, sum * 10 + 7);
}
int main()
{
    num = 0;
    for(int i = 2; i <= 18; i += 2) {
        DFS(1, 0, 0, i, 0);
    }sort(ans, ans + num);
    int t; scanf("%d", &t);
    while(t--) {
        __int64 n; scanf("%I64d", &n);
        int pos = lower_bound(ans, ans + num, n) - ans;
        if(pos != num) {
            printf("%I64d\n", ans[pos]);
        }
        else {
            printf("44444444447777777777\n");
        }
    }
    return 0;
}