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(≤n≤) cases

For each cases:

The only line contains a positive integer n(≤n≤). This number doesn't have leading zeroes.

Output

For each cases
Output the answer

Sample Input

Sample Output

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

 void dfs(ll sum,int c1,int c2){
    if(c1>=k1 && c2>=k2){
       ans[num++]=sum;
       return;
    }
    if(c1<k1) dfs(sum*+,c1+,c2);
    if(c2<k2) dfs(sum*+,c1,c2+);
 }

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

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：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 #define ll long long
 #define N 1000000
 int k1,k2;
 ll ans[N];
 int num;
 ll n;
 void dfs(ll sum,int c1,int c2){
    if(c1>=k1 && c2>=k2){
       ans[num++]=sum;
       return;
    }
    if(c1<k1) dfs(sum*+,c1+,c2);
    if(c2<k2) dfs(sum*+,c1,c2+);
 }
 void init(){
    num=;
    for(int i=;i<=;i+=){
       k1=k2=i/;
       dfs(,,);
    }
 }
 int main()
 {
     init();
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%I64d",&n);
         int cnt = lower_bound(ans,ans+num,n)-ans;
         if(cnt<num){
            printf("%I64d\n",ans[cnt]);
         }else{
            printf("44444444447777777777\n");
         }
     }
     return ;
 }