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

题目传送门

题目大意

　　求一个满足$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil - \sum_{i = 1}^{n} a_{i} \leqslant K$的最大正整数$d$。

　　整理一下可以得到条件是$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil \leqslant K + \sum_{i = 1}^{n} a_{i}$

　　有注意到$\left \lceil \frac{a_i}{d} \right \rceil$的取值个数不会超过$2\left \lceil \sqrt{a_i} \right \rceil$。

　　证明考虑对于$1 \leqslant d \leqslant \left \lceil \sqrt{a_i} \right \rceil$，至多根号种取值，当$d >\left \lceil \sqrt{a_i} \right \rceil$的时候，取值至多为$1, 2, \cdots, \left \lceil \sqrt{a_i} \right \rceil$，所以总共不会超过$2\left \lceil \sqrt{a_i} \right \rceil$个取值。

　　所以我们把所有$\left \lceil \frac{a_i}{d} \right \rceil$的取值当成一个点，安插在数轴上，排个序，就愉快地找到了所有分段了。因为每一段内的$d$都是等价的，所以只需要用每一段的左端点计算和，然后判断$\left \lfloor \frac{K + \sum_{i = 1}^{n} a_{i}}{\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil} \right \rfloor$是否在区间内，如果是就用它去更新答案。

Code

 /**

  * Codeforces

  * Problem#831F

  * Accepted

  * Time:997ms

  * Memory:100400k

  */

 #include <iostream>

 #include <cstdio>

 #include <ctime>

 #include <cmath>

 #include <cctype>

 #include <cstring>

 #include <cstdlib>

 #include <fstream>

 #include <sstream>

 #include <algorithm>

 #include <map>

 #include <set>

 #include <stack>

 #include <queue>

 #include <vector>

 #include <stack>

 #ifndef WIN32

 #define Auto "%lld"

 #else

 #define Auto "%I64d"

 #endif

 using namespace std;

 typedef bool boolean;

 const signed int inf = (signed)((1u << ) - );

 const signed long long llf = (signed long long)((1ull << ) - );

 const double eps = 1e-;

 const int binary_limit = ;

 #define smin(a, b) a = min(a, b)

 #define smax(a, b) a = max(a, b)

 #define max3(a, b, c) max(a, max(b, c))

 #define min3(a, b, c) min(a, min(b, c))

 template<typename T>

 inline boolean readInteger(T& u){

     char x;

     int aFlag = ;

     while(!isdigit((x = getchar())) && x != '-' && x != -);

     if(x == -) {

         ungetc(x, stdin);

         return false;

     }

     if(x == '-'){

         x = getchar();

         aFlag = -;

     }

     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');

     ungetc(x, stdin);

     u *= aFlag;

     return true;

 }

 #define LL long long

 int n;

 LL C;

 int* a;

 vector<LL> seg;

 template<typename T>

 T ceil(T a, T b) {

     return (a + b - ) / b;

 }

 inline void init() {

     readInteger(n);

     readInteger(C);

     seg.push_back();

     a = new int[(n + )];

     for(int i = , x; i <= n; i++) {

         readInteger(a[i]);

         for(int j = ; j * j <= a[i]; j++)

             seg.push_back(j), seg.push_back(ceil(a[i], j));

         C += a[i];

     }

     seg.push_back(llf);

 }

 LL res = ;

 inline void solve() {

     sort(seg.begin(), seg.end());

     int m = unique(seg.begin(), seg.end()) - seg.begin() - ;

     for(int i = ; i < m; i++) {

         LL l = seg[i], r = seg[i + ], temp = ;

         for(int i = ; i <= n; i++)

             temp += ceil((LL)a[i], l);

         LL d = C / temp;

         if(d >= l && d < r && d > res)

             res = d;

     }

     printf(Auto"\n", res);

 }

 int main() {

     init();

     solve();

     return ;

 }

Brute force

　　然后我们来讲点神仙做法。 orz orz orz.....

　　不妨设$C = K + \sum_{i = 1}^{n} a_{i}$

　　因为所有$\left \lceil \frac{a_i}{d} \right \rceil \geqslant 1$，所以$d\leqslant \left \lfloor \frac{C}{n} \right \rfloor$

　　设$d_0 = \left \lfloor \frac{C}{n} \right \rfloor$。

　　假装已经顺利地求出了$d_0, d_1, d_2, \cdots, d_k$，我们找到最大的$d_{k + 1}$满足：

$d_{k + 1}\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_k} \right \rceil \leqslant C$

　　当$d_{k + 1} = d_k$的时候我们就找到最优解了。

　　感觉很玄学？那我来证明一下。

定理1 $d_{k + 1} \leqslant d_{k}$

　　证明

当$k = 0$的时候，因为$\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_0} \right \rceil \geqslant n$，所以$d_{1} = \left \lfloor \frac{C}{\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_0} \right \rceil} \right \rfloor \leqslant \left \lfloor\frac{C}{n}\right \rfloor = d_0$。
当$k > 0$的时候，假设当$k = m - 1, (m \geqslant 0)$时成立，考虑当$k = m$的时候，由$d_{k} \leqslant d_{k - 1}$可得$\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_{k - 1}} \right \rceil \leqslant \sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_{k}} \right \rceil $，然后可得$d_{k + 1} \leqslant d_k$。

定理2 当$d_{k + 1} = d_{k}$，$d_k$是最优解

　　证明假设存在$d > d_k$满足条件

显然$d \leqslant d_0$。（不然直接不合法）
显然$d \neq d_j\ \ (0 \leqslant j < k)$。
假设$d_{j} < d < d_{j - 1}\ \ (0 < j \leqslant k)$，那么$C \geqslant d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil \geqslant d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d_{j}} \right \rceil $
由迭代做法可知$d\leqslant d_{j + 1}\leqslant d_j$，矛盾。

　　显然$d = d_k$时是合法的，所以$d_k$是最优解。

　　时间复杂度感觉很低，但是我只会证它不超过$O(n\sqrt{\frac{C}{n}})$

Code

 /**

  * Codeforces

  * Problem#831F

  * Accepted

  * Time: 31ms

  * Memory: 0k

  */

 #include <iostream>

 #include <cassert>

 #include <cstdlib>

 #include <cstdio>

 #include <vector>

 #include <queue>

 #ifndef WIN32

 #define Auto "%lld"

 #else

 #define Auto "%I64d"

 #endif

 using namespace std;

 typedef bool boolean;

 #define ll long long

 const int N = ;

 int n;

 ll C;

 int ar[N];

 inline void init() {

     scanf("%d"Auto, &n, &C);

     for (int i = ; i < n; i++)

         scanf("%d", ar + i), C += ar[i];

 }

 ll ceil(ll a, ll b) {

     return (a + b - ) / b;

 }

 inline void solve() {

     ll dcur = C / n, dans;

     do {

         swap(dcur, dans), dcur = ;

         for (int i = ; i < n; i++)

             dcur += ceil(ar[i], dans);

         dcur = C / dcur;

     } while (dcur != dans);

     printf(Auto"\n", dans);

 }

 int main() {

     init();

     solve();

     return ;

 }

更新日志

2018-1-28 补上jmr的做法
2018-10-22 给出证明