A. Banana

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
const int N = 1e5 + 10;
struct Node {
	int k;
	int cur;
	char c;
	int st () {
		return (k - 1) / cur + 1;
	}
}t[N];
int n, m;
string s;
bool operator< (const Node&a, const Node&b) {
	return ((a.k - 1) / a.cur + 1) < ((b.k - 1) / b.cur + 1);
}
void solve() {
	cin >> s;
	cin >> n;
	int cnt[26] = {0};
	for (auto t : s)
		cnt[t - 'a'] ++;
	string ans = "";
	priority_queue<Node> heap;
	for (int i = 0; i < 26; i ++)
		if (cnt[i])
		{
			heap.push({cnt[i], 1, i + 'a'});
			ans += (i + 'a');
		}
	if (n < heap.size()) {
		puts("-1");
		return;
	}
	n -= heap.size();
	while (n) {
		auto t = heap.top(); heap.pop();
		t.cur ++;
		ans += t.c;
		heap.push(t);
		n --;
	}
	auto t = heap.top();
	cout << t.st() << endl;
	cout << ans << endl;
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意:问至少拿多少个次以及构造一个可能拿的字符构成的字符串

题解:首先每个字母拿一次，然后还剩(k - 1) / a + 1因为已经拿了一次，所以加1，然后再判断是否合法，接着判断还需要拿多少个。按照紧急程排序