C - Lucky Numbers (easy)

Problem description

Petya 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 Petya came across a positive integer n. Help him to find the least super lucky number which is not less than n.

Input

The only line contains a positive integer n (1 ≤ n ≤ 10⁹). This number doesn't have leading zeroes.

Output

Output the least super lucky number that is more than or equal to n.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

4500

47

Output

4747

47

解题思路：题目要求输出不小于n的最小整数，这个整数要求只含有4和7这两个数字，并且4出现的次数等于7出现的次数。暴力打表（大概2分钟），水过！
打表代码：

 #include<iostream>
 using namespace std;
 typedef long long LL;
 bool judge(LL x){
     int t1 = , t2 = ;
     while (x){
         int a = x % ;
         if (a !=  && a != )return false;
         if (a == )t1++;
         if (a == )t2++;
         x /= ;
     }
     if (t1 == t2)return true;
     else return false;
 }
 int main(){
     for (LL i = ; i < ; ++i)
         if (judge(i)){ cout << i << ','; }
     return ;
 }

AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 LL n,obj[]={,,,,,,,,
         ,,,,,,,
         ,,,,,,,
         ,,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,,,,
         ,,,};
 int main(){
     cin>>n;
     for(int i=;;++i)
         if(obj[i]>=n){cout<<obj[i]<<endl;break;}
     return ;
 }

再贴一种很好理解的递归写法（反正我没想到QAQ，值得学习一下）AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 LL n,ans=1LL <<;
 void f(LL x,int y,int z){//通过x构造答案，y是4的个数，z是7点个数
     if(x>=n && y==z)ans=min(ans,x);//答案ans满足要求:1.大于等于n  2.只存在7和4，且7的个数等于4的个数
     if(x>n*)return;//答案肯定在n和n*100之间，所以x>n*100的时候退出；
     f(x*+,y+,z);//当前答案x加一位数4
     f(x*+,y,z+);//当前答案x加一位数7
 }
 int main(){
     cin>>n;
     f(,,);
     cout<<ans<<endl;
     return ;
 }