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

题目传送门

　　传送门I

　　传送门II

　　传送门III

题目大意

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

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

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

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

　　所以我们把所有$left lceil frac{a_i}{d} ight ceil$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$left lfloor frac{K + sum_{i = 1}^{n} a_{i}}{sum_{i = 1}^{n} left lceil frac{a_i}{d} ight ceil} ight floor$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 << 31) - 1);
const signed long long llf = (signed long long)((1ull << 63) - 1);
const double eps = 1e-6;
const int binary_limit = 128;
#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 = 1;
    while(!isdigit((x = getchar())) && x != '-' && x != -1);
    if(x == -1) {
        ungetc(x, stdin);    
        return false;
    }
    if(x == '-'){
        x = getchar();
        aFlag = -1;
    }
    for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
    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 - 1) / b;
}

inline void init() {
    readInteger(n);
    readInteger(C);
    seg.push_back(1);
    a = new int[(n + 1)];
    for(int i = 1, x; i <= n; i++) {
        readInteger(a[i]);
        for(int j = 1; 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 = 1;
inline void solve() {
    sort(seg.begin(), seg.end());
    int m = unique(seg.begin(), seg.end()) - seg.begin() - 1;
    for(int i = 0; i < m; i++) {
        LL l = seg[i], r = seg[i + 1], temp = 0;
        for(int i = 1; i <= n; i++)
            temp += ceil((LL)a[i], l);
        LL d = C / temp;
        if(d >= l && d < r && d > res)
            res = d;
    }
    printf(Auto"
", res);
}

int main() {
    init();
    solve();
    return 0;
}

　　然后我们来讲点神仙做法。142857 orz orz orz.....

　　不妨设$C = K + sum_{i = 1}^{n} a_{i}$

　　因为所有$left lceil frac{a_i}{d} ight ceil geqslant 1$，所以$dleqslant left lfloor frac{C}{n} ight floor$

　　设$d_0 =  left lfloor frac{C}{n} ight floor$。

　　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：

$d_{k + 1}sum_{i = 1}^{n} left lceil frac{a_i}{d_k} ight ceil leqslant  C$

　　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$sum_{i = 1}^{n} left lceil frac{a_i}{d_0} ight ceil geqslant n$，所以$d_{1} = left lfloor frac{C}{sum_{i = 1}^{n} left lceil frac{a_i}{d_0} ight ceil} ight floor leqslant left lfloorfrac{C}{n} ight floor = d_0$。
• 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}} ight ceil leqslant sum_{i = 1}^{n} left lceil frac{a_i}{d_{k}} ight ceil $，然后可得$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$d eq d_j   (0 leqslant j < k)$。
• 假设$d_{j} < d < d_{j - 1} (0 < j leqslant k)$，那么$C geqslant dsum_{i = 1}^{n} left lceil frac{a_i}{d} ight ceil geqslant dsum_{i = 1}^{n} left lceil frac{a_i}{d_{j}} ight ceil $
  由迭代做法可知$dleqslant d_{j + 1}leqslant d_j$，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$O(nsqrt{frac{C}{n}})$

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 = 105;

int n;
ll C;
int ar[N];

inline void init() {
    scanf("%d"Auto, &n, &C);
    for (int i = 0; i < n; i++)
        scanf("%d", ar + i), C += ar[i];
}

ll ceil(ll a, ll b) {
    return (a + b - 1) / b;
}

inline void solve() {
    ll dcur = C / n, dans;
    do {
        swap(dcur, dans), dcur = 0;
        for (int i = 0; i < n; i++)
            dcur += ceil(ar[i], dans);
        dcur = C / dcur;
    } while (dcur != dans);
    printf(Auto"
", dans);
}

int main() {
    init();
    solve();
    return 0;
}

更新日志

• 2018-1-28 补上jmr的做法
• 2018-10-22 给出证明