C - Lucky Numbers (easy)

Posted 2020-11-19 acgoto

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

Output

4747

Input

47

Output

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：

 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代码：

 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 }

 再贴另一种很好理解的递归写法（反正我没想到QAQ，不过代码思路真的很溜，值得学习一下，涨一下见识）AC代码：

 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){//通过x构造答案，y是4的个数，z是7点个数
 6     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
 7     if(x>n*100)return;//答案肯定在n和n*100之间，所以x>n*100的时候退出；
 8     f(x*10+4,y+1,z);//当前答案x加一位数4
 9     f(x*10+7,y,z+1);//当前答案x加一位数7
10 }
11 int main(){
12     cin>>n;
13     f(0,0,0);
14     cout<<ans<<endl;
15     return 0;
16 }