Vladimir wants to modernize partitions in his office. To make the office more comfortable he decided to remove a partition and plant several bamboos in a row. He thinks it would be nice if there are n bamboos in a row, and the i-th from the left is a_i meters high.

Vladimir has just planted n

Vladimir wants to check the bamboos each d

What is the maximum value d he can choose so that he can achieve what he wants without cutting off more than k

The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 10¹¹) — the number of bamboos and the maximum total length of cut parts, in meters.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the required heights of bamboos, in meters.

Print a single integer — the maximum value of d such that Vladimir can reach his goal.

3 4
1 3 5

3

3 40
10 30 50

32

In the first example Vladimir can check bamboos each 3 days. Then he will cut the first and the second bamboos after 3 days, and the third bamboo after 6 days. The total length of cut parts is 2 + 0 + 1 = 3 meters.

　　題目大意 求這樣一個最大的d使得

　　整理一下可以得到條件是

　　有註意到的取值個數為。

　　還是說明一下，設。

　　1)當時，顯然至多個取值。

　　2)當時，顯然有，這個的取值個數也是的。

　　因此可以發現當d等於一個值的時候可能和另一個d值的時候是等價的。

　　所以我們把所有的取值當成一個點，安插在數軸上，然後一段一段地進行枚舉。因為每一段內的d都是等價的，所以只需要用每一段的左端點計算和，然後判斷是否在區間內，如果是就用它去更新答案。

Code

  1 /**
  2  * Codeforces
  3  * Problem#831F
  4  * Accepted
  5  * Time:997ms
  6  * Memory:100400k
  7  */
  8 #include <iostream>
  9 #include <cstdio>
 10 #include <ctime>
 11 #include <cmath>
 12 #include <cctype>
 13 #include <cstring>
 14 #include <cstdlib>
 15 #include <fstream>
 16 #include <sstream>
 17 #include <algorithm>
 18 #include <map>
 19 #include <set>
 20 #include <stack>
 21 #include <queue>
 22 #include <vector>
 23 #include <stack>
 24 #ifndef WIN32
 25 #define Auto "%lld"
 26 #else
 27 #define Auto "%I64d"
 28 #endif
 29 using namespace std;
 30 typedef bool boolean;
 31 const signed int inf = (signed)((1u << 31) - 1);
 32 const signed long long llf = (signed long long)((1ull << 63) - 1);
 33 const double eps = 1e-6;
 34 const int binary_limit = 128;
 35 #define smin(a, b) a = min(a, b)
 36 #define smax(a, b) a = max(a, b)
 37 #define max3(a, b, c) max(a, max(b, c))
 38 #define min3(a, b, c) min(a, min(b, c))
 39 template<typename T>
 40 inline boolean readInteger(T& u){
 41     char x;
 42     int aFlag = 1;
 43     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
 44     if(x == -1) {
 45         ungetc(x, stdin);    
 46         return false;
 47     }
 48     if(x == ‘-‘){
 49         x = getchar();
 50         aFlag = -1;
 51     }
 52     for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
 53     ungetc(x, stdin);
 54     u *= aFlag;
 55     return true;
 56 }
 57 
 58 #define LL long long
 59 
 60 int n;
 61 LL C;
 62 int* a;
 63 vector<LL> seg;
 64 
 65 template<typename T>
 66 T ceil(T a, T b) {
 67     return (a + b - 1) / b;
 68 }
 69 
 70 inline void init() {
 71     readInteger(n);
 72     readInteger(C);
 73     seg.push_back(1);
 74     a = new int[(n + 1)];
 75     for(int i = 1, x; i <= n; i++) {
 76         readInteger(a[i]);
 77         for(int j = 1; j * j <= a[i]; j++)
 78             seg.push_back(j), seg.push_back(ceil(a[i], j));
 79         C += a[i];
 80     }
 81     seg.push_back(llf);
 82 }
 83 
 84 LL res = 1;
 85 inline void solve() {
 86     sort(seg.begin(), seg.end());
 87     int m = unique(seg.begin(), seg.end()) - seg.begin() - 1;
 88     for(int i = 0; i < m; i++) {
 89         LL l = seg[i], r = seg[i + 1], temp = 0;
 90         for(int i = 1; i <= n; i++)
 91             temp += ceil((LL)a[i], l);
 92         LL d = C / temp;
 93         if(d >= l && d < r && d > res)
 94             res = d;
 95     }
 96     printf(Auto"\n", res);
 97 }
 98 
 99 int main() {
100     init();
101     solve();
102     return 0;
103 }

Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem F (Codeforces 831F) - 數論 - 暴力