hdu5676 ztr loves lucky numbers DFS+二分

最新推荐文章于 2021-10-21 18:26:48 发布

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本文讨论了正整数中的幸运数和超级幸运数的概念，并提供了寻找至少大于给定数字n的最小超级幸运数的方法。通过数位深度优先搜索（DFS）找到所有超级幸运数，然后使用二分查找定位目标数。

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 330 Accepted Submission(s): 133

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

当n > 777777777444444444结果会爆long long，需要对这一个特判一下，别的都数位dfs

然后将dfs找到的数都放到一个数组a中，排序后用二分找大于等于n的第一个数

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
typedef long long ll;

ll a[266666];
int len;

void dfs(int x, int y, ll num) {
	if (x == 0 && y == 0) {
		a[++len] = num;
		return;
	}
	if (x > 0)
		dfs(x - 1, y, num * 10 + 4);
	if (y > 0)
		dfs(x, y - 1, num * 10 + 7);
}

int main()
{
	int T;
	ll n;
	scanf("%d", &T);
	len = 0;
	for (int i = 2; i <= 18; i += 2) {
		dfs(i / 2, i / 2, 0);
	}
	sort(a + 1, a + 1 + len);
	while (T--) {
		scanf("%I64d", &n);
		if (n > 777777777444444444) {
			puts("44444444447777777777");
		}
		else {
			int l = 1, r = len;
			ll ans;
			while (l <= r) {
				int mid = (l + r) / 2;
				if (a[mid] >= n) {
					ans = a[mid];
					r = mid - 1;
				}
				else {
					l = mid + 1;
				}
			}
			printf("%I64d\n", ans);
		}
	}
	return 0;
}