結果
| 問題 |
No.78 クジ付きアイスバー
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2024-06-28 14:54:13 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 2 ms / 5,000 ms |
| コード長 | 2,783 bytes |
| コンパイル時間 | 1,368 ms |
| コンパイル使用メモリ | 163,584 KB |
| 実行使用メモリ | 6,820 KB |
| 最終ジャッジ日時 | 2025-01-01 18:05:07 |
| 合計ジャッジ時間 | 2,851 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 35 |
ソースコード
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep_r(i,a,b) for(int i=a;i>=b;i--)
#define each(a,x) for (auto& x : a)
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define sz(x) int(x.size())
#define so(x) sort(all(x))
#define so_r(x) sort(all(x),greater<int>())
#define lb lower_bound
#define ub upper_bound
const char nl = '\n';
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int bit_cnt(int x){
return __builtin_popcount(x);
}
ll bex(ll a, ll b, ll mod = 1e9 + 7){ll res = 1LL; while(b){ if (b&1) res = res * a % mod; a = a * a % mod; b >>= 1;} return res;}
template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}
const int MOD = 998244353;
const int N = 5e5 + 5;
const ll INF = 1e16;
int n;
ll k;
string s;
// https://yukicoder.me/problems/no/78
// brief: n icecreams in 1 box, 3 types of icecream: "truot", "trung 1 que", "trung 2 que"
// lay lan luot que kem theo thu tu, type cua que kem duoc ghi trong S (0 || 1 || 2)
// can mua it nhat bao nhieu que kem de co the an duoc it nhat k que kem
// n <= 50, k <= 2e9
pair<int,int> one_box(){
int buy = 0;
int remain = 0;
rep(i,1,n){
buy++;
int j = i;
int prize = s[i-1] - '0';
while (prize && j < n) {
prize--;
j++;
if (j > i) prize += s[j-1] - '0';
}
i = j;
remain = prize;
}
return {buy, remain};
}
int find1(int k){
int buy = 0;
if (k == 0) return 0;
rep(i,1,n){
buy++;
int j = i;
int prize = s[i-1] - '0';
while (prize && j < n) {
prize--;
j++;
if (j > i) prize += s[j-1] - '0';
}
i = j;
if (i >= k) return buy;
}
return n;
}
void solve(){
cin >> n >> k >> s;
pair<int,int> p = one_box();
int buy = p.first, remain = p.second;
// cout << buy << ' ' << remain << nl;
if (k <= n) {
cout << find1(k) << nl;
return;
}
if (k > n && remain >= buy){
cout << buy << nl;
return;
}
// extra buy times we need to spend for full fill other box
int next_buy = buy - remain;
ll boxes = k / n; // total full-boxes we need to buy
ll still = k % n; // number of icecream still need to buy
ll ans = 1 * buy + next_buy * (boxes - 1) + max(find1(still) - remain, 0);
cout << ans << nl;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
// cin >> t;
while (t--){
solve();
}
}