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:
代码语言:javascript复制A BOutput
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
代码语言:javascript复制2 3Sample Output 1
代码语言:javascript复制6When 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
代码语言:javascript复制123 456Sample Output 2
代码语言:javascript复制18696Sample Input 3
代码语言:javascript复制100000 99999Sample Output 3
代码语言:javascript复制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
代码语言:javascript复制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
代码语言:javascript复制ll A, B;
//---------------------------------------------------------------------------------------------------
void _main() {
cin >> A >> B;
cout << lcm(A, B) << endl;
}


