hdu 5676 ztr loves lucky numbers（dfs+离线）

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

直接暴力显然TLE，考虑按位DFS

每一位只可能是4或7

所以根据这个来DFS即可，时间复杂度O(T*2^{log_{10}n})O(T∗2?log?10??n??)

考虑到T特别大，不可能每次都DFS

而经过计算，2^{18}=2621442?18??=262144，所以全部储存下来

对于每次询问，二分即可

考虑一个边界条件，即当结果爆ll怎么办？

即答案应当为10个4、10个7的时候，显然unsigned long long也不行

那么只能采用特判了

首先,由于每一位只能是4或7,且
2^{18}=2621442?18??=262144,故而考虑离线打表,即利用dfs求解出所有只含数字4和7,且4和7数量相等的数

1 void dfs(ll sum,int c1,int c2){
2    if(c1>=k1 && c2>=k2){
3       ans[num++]=sum;
4       return;
5    }
6    if(c1<k1) dfs(sum*10+4,c1+1,c2);
7    if(c2<k2) dfs(sum*10+7,c1,c2+1);
8 }

结果存储在ans数组中,此时可以得到符合条件的数有66196+1个,而数据有T组,好吧,那就二分查找

1 int cnt = lower_bound(ans,ans+num,n)-ans;

需要注意的一点是,当n>777777777444444444的时候,18位数里面已经没有比这更大的符合条件的数了,那么只能特判一下,解为最小的20位符合条件的数,即44444444447777777777

AC代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 #define ll long long
 6 #define N 1000000
 7 int k1,k2;
 8 ll ans[N];
 9 int num;
10 ll n;
11 void dfs(ll sum,int c1,int c2){
12    if(c1>=k1 && c2>=k2){
13       ans[num++]=sum;
14       return;
15    }
16    if(c1<k1) dfs(sum*10+4,c1+1,c2);
17    if(c2<k2) dfs(sum*10+7,c1,c2+1);
18 }
19 void init(){
20    num=0;
21    for(int i=2;i<=18;i+=2){
22       k1=k2=i/2;
23       dfs(0,0,0);
24    }
25 }
26 int main()
27 {
28     init();
29     int t;
30     scanf("%d",&t);
31     while(t--){
32         scanf("%I64d",&n);
33         int cnt = lower_bound(ans,ans+num,n)-ans;
34         if(cnt<num){
35            printf("%I64d\n",ans[cnt]);
36         }else{
37            printf("44444444447777777777\n");
38         }
39     }
40     return 0;
41 }