Codeforces Round #465 (Div. 2) A. Fafa and his Company 水题
2018-02-21 22:30
465 查看
A. Fafa and his Companytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputFafa owns a company that works on huge projects. There are n employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.Fafa finds doing this every time is very tiring for him. So, he decided to choose the best l employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.Given the number of employees n, find in how many ways Fafa could choose the number of team leaders l in such a way that it is possible to divide employees between them evenly.InputThe input consists of a single line containing a positive integer n (2 ≤ n ≤ 105) — the number of employees in Fafa's company.OutputPrint a single integer representing the answer to the problem.ExamplesinputCopy
choose 2 employees as team leaders with 4 employees under the responsibility of each of them.
choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
题意:工厂给定一种管理方式,使得每个领导的手下个数均相等且领导互相平级,管理无交叉。给定工人的个数,求有多少种分配方案。
将工人总数减去领导数,再看领导数是否可被剩余人数整除即可。
代码如下:#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
while (~scanf("%d",&n)){
int ans=0;
for (int i=1;i<=n/2;i++){
int rest=n-i;
if (rest%i==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}
2output
1inputCopy
10output
3NoteIn the second sample Fafa has 3 ways:choose only 1 employee as a team leader with 9 employees under his responsibility.
choose 2 employees as team leaders with 4 employees under the responsibility of each of them.
choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
题意:工厂给定一种管理方式,使得每个领导的手下个数均相等且领导互相平级,管理无交叉。给定工人的个数,求有多少种分配方案。
将工人总数减去领导数,再看领导数是否可被剩余人数整除即可。
代码如下:#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
while (~scanf("%d",&n)){
int ans=0;
for (int i=1;i<=n/2;i++){
int rest=n-i;
if (rest%i==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}
相关文章推荐
- Codeforces Round #465 (Div. 2) A. Fafa and his Company
- codeforces 935 A. Fafa and his Company
- Fafa and his Company
- A. Fafa and his Company
- A. Fafa and his Company
- cf 465 A. Fafa and his Company
- Codeforces Round #378 (Div. 2) A. Grasshopper And the String(水题求跳跃距离)
- Codeforces Round #304 (Div. 2) B. Soldier and Badges 水题
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) B. Bear and Blocks 水题
- Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) A. Neverending competitions 水题
- Codeforces Round #365 (Div. 2) -- C. Chris and Road (思路水题)
- 【Codeforces Round 330 (Div 2)A】【水题】Vitaly and Night 多少间房间有人
- Codeforces Round #465 (Div. 2)E. Fafa and Ancient Mathematics(CF935E)(树形动规)
- ACM 水题 Wallace and His Pet
- Codeforces Round #383 (Div. 2) A. Arpa’s hard exam and Mehrdad’s naive cheat(水题)
- Codeforces Round #384 (Div. 2) A. Vladik and flights【思维】水题
- Codeforces Round #378 (Div. 2)A.Grasshopper And the String【水题】模拟
- Codeforces Round #404 (Div. 2) B. Anton and Classes 水题
- 【Codeforces Round 354 (Div 2)A】【水题 贪心】Nicholas and Permutation 恰好交换一次的pos[n]-pos[1]
- Codeforces Round #304 (Div. 2) C. Soldier and Cards 水题