勤劳做法：

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long n,q[99]={47,74,4477,4747,4774,7447,7474,7744,444777,447477,447747,447774,474477,474747,474774,477447,477474,477744,744477,744747,744774,747447,747474,747744,774447,774474,774744,777444,44447777,44474777,44477477,44477747,44477774,44744777,44747477,44747747,44747774,44774477,44774747,44774774,44777447,44777474,44777744,47444777,47447477,47447747,47447774,47474477,47474747,47474774,47477447,47477474,47477744,47744477,47744747,47744774,47747447,47747474,47747744,47774447,47774474,47774744,47777444,74444777,74447477,74447747,74447774,74474477,74474747,74474774,74477447,74477474,74477744,74744477,74744747,74744774,74747447,74747474,74747744,74774447,74774474,74774744,74777444,77444477,77444747,77444774,77447447,77447474,77447744,77474447,77474474,77474744,77477444,77744447,77744474,77744744,77747444,77774444,4444477777};
    cin>>n;
    cout<<q[lower_bound(q,q+99,n)-q];
    return 0;
}

我用的是广搜做法： 1：先将4和7放入队列，依次在后面加4和7（如果要在x后加y,可以用x*10+y的方式）。 2：在添加数（代码中的now)的过程中，判断数是否满足条件（now >= n && lucky(now)),如果可以这个数就是答案，直接输出即可。lucky(now)表示判断now 4的位数和7的位数是否相等。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,a,b,c) for (int i = a;i <= b;i += c)
int n;
queue<int> q;
bool lucky(int x) {
	int n4=0,n7=0;
	while (x > 0) {
		if (x % 10 == 4) n4++;
		else if (x % 10 == 7) n7++;
		x /= 10;
	}
	return (n4 == n7);
}//判断now 4的位数和7的位数是否相等
void bfs() {
	q.push(4); q.push(7);
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		if (now >= n && lucky(now)) {
			cout << now;
			break;
		}
		q.push(now*10+4);
		q.push(now*10+7);
	}
}//广搜答案
signed main()
{
 	cin >> n;
	bfs();
    return 0;
}

666

给出深搜做法