hdoj ztr loves lucky numbers 5676 （dfs模拟）

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 918    Accepted Submission(s): 389

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)

问题描述

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

输入描述

有T(1\leq\;T\leq\;10^{5})T(1≤T≤10​5​​)组数据，每组数据一个正整数$n$，1\leq\;n\leq\;10^{18}1≤n≤10​18​​

输出描述

有$T$行，每行即答案

输入样例

2
4500
47

输出样例

4747
47

Hint

请尽可能地优化算法，考虑全面

//没啥说的，就是一个dfs模拟，但是得注意一个特判，见代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ll long long
#define N 100010
using namespace std;
ll a[N];
int cnt;
void dfs(int l,int r,int len,ll num)
{
	if(l>len/2||r>len/2) return ;
	if(l+r==len&&l==r)
	{
		a[cnt++]=num;
		return ;
	}
	if(l<=len/2) dfs(l+1,r,len,num*10+4);
	if(r<=len/2) dfs(l,r+1,len,num*10+7);
	
}
int main()
{
	cnt=0;
	for(int i=2;i<=18;i+=2)
		dfs(0,0,i,0);
	int t;
	ll n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		if(n>777777777444444444ll)   //特判 
			printf("44444444447777777777\n");
		else
			printf("%lld\n",a[lower_bound(a,a+cnt,n)-a]);
	}
	return 0;
}