Codeforces Round #674 C Increase and Copy【数学】

最新推荐文章于 2022-01-11 22:57:54 发布

Dousir9

最新推荐文章于 2022-01-11 22:57:54 发布

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分类专栏：数学二分文章标签：算法

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题目链接

题目描述：

在这里插入图片描述

题解：

it is pretty intuitive that we firstly need to do all increaments and only then copy numbers. You could notice that the answer does not exceed $n^{1/2}$ so we can iterate from 1 to $\lfloor n^{1/2} \rfloor$ and fix number we will copy. Let it be x. Then we need x - 1 moves to obtain it and also need $\lceil \frac{n-x}{x} \rceil$ moves to get the enough number of copies. So we can update the answer with this number of moves.

二分答案：

我比赛的时候用的二分答案做的，首先令 x 为increments 的次数，y 为 copy 的次数，n 为总 moves 的次数，有 n = x + y。设 w 为经过 x 次 increment 和 y 次 copy 后得到的值，即 w = (1 + x) * y，经过与 n = x + y 联立，求导并代入，最终得到 $(\frac{n+1}{2})^2$ 为 n 次操作的最大值，然后进行二分答案即可，最终得到的结果要减 1。

Tips：

If you want to calculate $\lceil \frac{X}{Y} \rceil$ then you do this: (X + Y - 1) / Y or (X - 1) / Y + 1 .
this works because division is rounded down if you cast the result to int

做法1(93ms)：

#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
//#include <unordered_set>
//#include <unordered_map>
#include <deque>
#include <list>
#include <iomanip>
#include <algorithm>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
//#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
//cout << fixed << setprecision(2);
//cout << setw(2);
const int N = 2e5 + 6, M = 1e9 + 7;



int main() {
    //freopen("/Users/xumingfei/Desktop/ACM/test.txt", "r", stdin);
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int ans = 1e9;
        for (int i = 1; i * i <= n; i++) {
            ans = min(ans, i - 1 + (n - i - 1) / i + 1);
        }
        cout << ans << '\n';
    }
    return 0;
}

二分答案(15ms)：

#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
//#include <unordered_set>
//#include <unordered_map>
#include <deque>
#include <list>
#include <iomanip>
#include <algorithm>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
//#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
//cout << fixed << setprecision(2);
//cout << setw(2);
const int N = 2e5 + 6, M = 1e9 + 7;



int main() {
    //freopen("/Users/xumingfei/Desktop/ACM/test.txt", "r", stdin);
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int l = 0, r = 1e9;
        while (l + 1 < r) {
            double mid = (l + r) >> 1;
            if (pow((mid + 1) / 2, 2) >= n) {
                r = mid;
            } else {
                l = mid;
            }
        }
        if (pow((1.0 * l + 1) / 2, 2) >= n) {
            cout << l - 1 << '\n';
        } else {
            cout << r - 1 << '\n';
        }
    }
    return 0;
}