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

4500

4747

47

47
解題思路：題目要求輸出不小於n的最小整數，這個整數要求只含有4和7這兩個數字，並且4出現的次數等於7出現的次數。暴力打表（大概2分鐘），水過！
打表代碼：

 1 #include<iostream>
 2
 
 using namespace std;
 3 typedef long long LL;
 4 bool judge(LL x){
 5     int t1 = 0, t2 = 0;
 6     while (x){
 7         int a = x % 10;
 8         if (a != 4 && a != 7)return false;
 9         if (a == 4)t1++;
10         if (a == 7)t2++;
11         x /= 10;
12     }
13     if (t1 == t2)return true;
14     else return false;
15 }
16 int main(){
17     for (LL i = 47; i < 10000000000; ++i)
18         if (judge(i)){ cout << i << ‘,‘; }
19     return 0;
20 }

AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL n,obj[]={47,74,4477,4747,4774,7447,7474,7744,
 5         444777,447477,447747,447774,474477,474747,474774,
 6         477447,477474,477744,744477,744747,744774,747447,
 7         747474,747744,774447,774474,774744,777444,44447777,
 8         44474777,44477477,44477747,44477774,44744777,44747477,
 9         44747747,44747774,44774477,44774747,44774774,44777447,
10         44777474,44777744,47444777,47447477,47447747,47447774,
11         47474477,47474747,47474774,47477447,47477474,47477744,
12         47744477,47744747,47744774,47747447,47747474,47747744,
13         47774447,47774474,47774744,47777444,74444777,74447477,
14         74447747,74447774,74474477,74474747,74474774,74477447,
15         74477474,74477744,74744477,74744747,74744774,74747447,
16         74747474,74747744,74774447,74774474,74774744,74777444,
17         77444477,77444747,77444774,77447447,77447474,77447744,
18         77474447,77474474,77474744,77477444,77744447,77744474,
19         77744744,77747444,77774444,4444477777};
20 int main(){
21     cin>>n;
22     for(int i=0;;++i)
23         if(obj[i]>=n){cout<<obj[i]<<endl;break;}
24     return 0;
25 }

再貼另一種很好理解的遞歸寫法（反正我沒想到QAQ，不過代碼思路真的很溜，值得學習一下，漲一下見識）AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL n,ans=1LL <<60;
 5 void f(LL x,int y,int z){//通過x構造答案，y是4的個數，z是7點個數
 6     if(x>=n && y==z)ans=min(ans,x);//答案ans滿足要求:1.大於等於n  2.只存在7和4，且7的個數等於4的個數
 7     if(x>n*100)return;//答案肯定在n和n*100之間，所以x>n*100的時候退出；
 8     f(x*10+4,y+1,z);//當前答案x加一位數4
 9     f(x*10+7,y,z+1);//當前答案x加一位數7
10 }
11 int main(){
12     cin>>n;
13     f(0,0,0);
14     cout<<ans<<endl;
15     return 0;
16 }

C - Lucky Numbers (easy)