秒杀AtCode ABC148 - C - Snack

AtCode ABC148 - C - Snack

标签

• 数学、最大公约数、最小公倍数

唯一了解完题意，几分钟做出来的题，不要觉得水平提高了，是这道题太水了，想明白了，分分钟搞定！

题目地址

C - Snack

• https://atcoder.jp/contests/abc148/tasks/abc148_c?lang=en

问题描述

Problem Statement

Takahashi is organizing a party.

At the party, each guest will receive one or more snack pieces.

Takahashi predicts that the number of guests at this party will be A or B.

Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.

We assume that a piece cannot be divided and distributed to multiple guests.

Constraints

• 1 \leq A, B \leq 10^5
• A \neq B
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.

Sample Input 1

2 3

Sample Output 1

6

When we have six snack pieces, each guest can take three pieces if we have two guests, and each guest can take two if we have three guests.

Sample Input 2

123 456

Sample Output 2

18696

Sample Input 3

100000 99999

Sample Output 3

9999900000

题意

• 高桥想举办一次聚会
  • 参加的人每个人都会分配1个以上的点心
  • 求不管A个人或者B个人都可以等分配的点心个数的最小值
  • 1个点心不能拆开分

思路

• 求最小公倍数
  • 这道题也是我刷题以来做的最快的一道C级题，题目比较水

题解

小码匠

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long a, b;
    cin >> a >> b;
    cout << a * b / __gcd(a, b);
}

官方题解

ll A, B;
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> A >> B;
    cout << lcm(A, B) << endl;
}