HDU 5676 ztr loves lucky numbers dfs+二分

ztr loves lucky numbers

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                      Total Submission(s): 938    Accepted Submission(s): 400

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)

 

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点击打开题目链接

dfs 出 long long 以内的所有幸运数，二分查找，超过 long long 的特判

#include <cstdio> 
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

typedef long long LL;
const int maxn = 300000;
LL num[maxn];
int cnt;

void dfs(LL sum, int sf, int ss)	//当前和，4的使用个数，7的使用个数
{
	if (sf + ss > 18) return;		//超过18位，不再搜索了 
	if (sf == ss) num[cnt++] = sum;	//4和7的个数相同，保存 
	
	dfs(10 * sum + 4, sf + 1, ss);	//当前数后面添加一个4 
	dfs(10 * sum + 7, sf, ss + 1);	//添加一个7 
}

int main()
{
	cnt = 0;
	dfs(0, 0, 0);
	sort(num, num + cnt);
	int t;
	LL n;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lld", &n);
		int L = 1, R = cnt;		//二分查找 
		while (L < R)
		{
			int M = L + (R - L) / 2;
			if (num[M] < n) L = M + 1;
			else R = M;
		}
		if (L == cnt) printf("44444444447777777777\n");	//超LL，特判 
		else printf("%lld\n", num[L]);
	}
	return 0;
}

可以用C++的STL中lower_bound 直接二分查找

#include <cstdio> 
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

typedef long long LL;
const int maxn = 300000;
LL num[maxn];
int cnt;

void dfs(LL sum, int sf, int ss)	//当前和，4的使用个数，7的使用个数
{
	if (sf + ss > 18) return;		//超过18位，不再搜索了 
	if (sf == ss) num[cnt++] = sum;	//4和7的个数相同，保存 
	
	dfs(10 * sum + 4, sf + 1, ss);	//当前数后面添加一个4 
	dfs(10 * sum + 7, sf, ss + 1);	//添加一个7 
}

int main()
{
	cnt = 0;
	dfs(0, 0, 0);
	sort(num, num + cnt);
	int t;
	LL n;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lld", &n);
		if (n == 0) 			//题中说的n>=1,可是不加就WA了 
		{
			printf("47\n");
			continue;
		}
		int k = lower_bound(num, num + cnt, n) - num; 	//二分查找 
	
		if (k == cnt) printf("44444444447777777777\n");	//超LL，特判 
		else printf("%lld\n", num[k]);
	}
	return 0;
}