B. A Balanced Problemset

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-27

37.B. A Balanced Problemset2024-01-28

38.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

忠告1：要学会计算时间复杂度！！

忠告2：要学会抓事实，不要掉进题目直观模拟的陷阱里

事实

1.任意k个数的 $gcd$ 一定可以是这k个数的最小值,这里以 $k=3$ 举例
假设 $gcd(a_1,a_2,a_3)=m$ ,则 $a_1=k_{1}m,a_2=k_{2}m,a_3=k_{3}m$,其中 $k_1,k_2,k_3$ 都是整数
那么可以通过移项变成 $a_1=(k_1+k_3-1)m,a_2=k_{2}m,a_3=m$
2.将n分成k个数的和，这k个数中的最小值 $a_{min}\le \frac{n}{k}$ 对任何n,k都成立
很显然啊，拿 $k=2$ 的情况举个例子就很明显了，$k$ 推广开来也是一样的
3.根据事实12，当 $a_{min}$ 是 $n$ 的因子时，答案就是 $a_{min}$
于是我们可以 $putitintosimplewords:$ ，找出 $n$ 的小于 $\frac{n}{k}$ 的 最大因子

判断时间复杂度

最坏情况： $t={10}^3,n={10}^8,k=1$ ， ${10}^{11}>{10}^8$
直接从 $\frac{n}{k}$ 开始往下遍历很大概率T
我们考虑优化
由于答案一定是 $n$ 的因子，而因子成对出现(包括1和自身),所以我们开根号找因子，每找一个因子相当于找到两个因子，对每个因子进行判断是否小于 $\frac{n}{k}$
此时 最坏情况 ${10}^7$

$Code$

#include<bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false); 
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t--) {
        int n, k;
        cin >> n >> k;
        int result = 1; 
        for(int i = 1; i <= sqrt(n); i++) {
            if(n % i == 0) {
                if(i <= n / k) result = max(result, i);
                if(n / i <= n / k) result = max(result, n / i);
            }
        }
        cout << result << endl;
    }
    return 0;
}