【小码匠自习室】ABC144-C：一个字：爽；两个字：好爽；三个字：特别爽

碎碎念

• 一个字：爽；两个字：好爽；三个字：特别爽。这是迄今为止做的最爽的一道题。
  • 老码农：爽你个大头鬼，老做爽题，你的水平就会停滞不前

ABC144-C-01

标签

• 数学、约数分解

题目地址

• C - Walk on Multiplication Table
  • https://atcoder.jp/contests/abc144/tasks/abc144_c

题目描述

Problem Statement

Takahashi is standing on a multiplication table with infinitely many rows and columns.

The square (i,j) contains the integer i

j. Initially, Takahashi is standing at (1,1).

In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).

Given an integer N, find the minimum number of moves needed to reach a square that contains N.

Constraints

• 2 \leq N \leq 10^{12}
• N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum number of moves needed to reach a square that contains the integer N.

Sample Input 1

10

Sample Output 1

5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.

Sample Input 2

50

Sample Output 2

13

(5, 10) can be reached in 13 moves.

Sample Input 3

10000000019

Sample Output 3

10000000018

Both input and output may be enormous.

题解

小码匠题解

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long n;
    cin >> n;
    for(long long i = sqrt(n); i > 0; --i) {
        if(n % i == 0) {
            cout << i - 1 + (n / i) - 1;
            return;
        }
    }
}

参考题解

#include <bits/stdc++.h>
using namespace std;

int main() {
    long long N;
    cin >> N;
    long long res = N + 1 - 2; // (a, b) = (1, N)
    for (long long a = 1; a * a <= N; ++a) {
        if (N % a == 0) res = min(res, a + N/a - 2);
    }
    cout << res << endl;
}