A. Vasya and Socks
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every m-th day (at days with numbers m, 2m, 3m, ...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?

Input

The single line contains two integers n and m (1 ≤ n ≤ 100; 2 ≤ m ≤ 100), separated by a space.

Output

Print a single integer — the answer to the problem.

Examples
input
2 2
output
3
input
9 3
output
13
Note

In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.

In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.

【代码】:

#include <bits/stdc++.h>

using namespace std;
int sum;
int main()
{
int n,m;
while(cin>>n>>m)
{
for(int i=n,j=;i>=;i--,j++)//i:袜子个数
{
if(j%m==)
i++;
sum++;
}
cout<<sum<<endl;
}
return ;
}

双指针

#include <bits/stdc++.h>

using namespace std;

int main()
{
int n,m;
while(cin>>n>>m)
{
int day=;
int k=;
while(n--)//袜子消耗
{
if(day%m==)
{
n++;//当是m的倍数的时候再购进一双袜子
}
day++;//不管有没有袜子穿日子总要过下去
k++;//维持的日子
}
printf("%d\n",k);
}
return ;
}

多变量记录

Codeforces Round #262 (Div. 2) A. Vasya and Socks【暴力/模拟/袜子在可以在合法情况下增加后用几天】的更多相关文章

  1. codeforces水题100道 第十五题 Codeforces Round #262 (Div. 2) A. Vasya and Socks (brute force)

    题目链接:http://www.codeforces.com/problemset/problem/460/A题意:Vasya每天用掉一双袜子,她妈妈每m天给他送一双袜子,Vasya一开始有n双袜子, ...

  2. Codeforces Round #262 (Div. 2)460A. Vasya and Socks(简单数学题)

    题目链接:http://codeforces.com/contest/460/problem/A A. Vasya and Socks time limit per test 1 second mem ...

  3. Codeforces Round #281 (Div. 2) A. Vasya and Football 暴力

    A. Vasya and Football   Vasya has started watching football games. He has learned that for some foul ...

  4. Codeforces Round #372 (Div. 2) A B C 水 暴力/模拟 构造

    A. Crazy Computer time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  5. Codeforces Round #262 (Div. 2) 1003

    Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 2 ...

  6. Codeforces Round #262 (Div. 2) 1004

    Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory lim ...

  7. 构造水题 Codeforces Round #206 (Div. 2) A. Vasya and Digital Root

    题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algori ...

  8. Codeforces Round #297 (Div. 2)D. Arthur and Walls 暴力搜索

    Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx ...

  9. Codeforces Round #262 (Div. 2)解题报告

    详见:http://robotcator.logdown.com/posts/221514-codeforces-round-262-div-2 1:A. Vasya and Socks   http ...

随机推荐

  1. 《Cracking the Coding Interview》——第18章:难题——题目12

    2014-04-29 04:36 题目:最大子数组和的二位扩展:最大子矩阵和. 解法:一个维度上进行枚举,复杂度O(n^2):另一个维度执行最大子数组和算法,复杂度O(n).总体时间复杂度为O(n^3 ...

  2. 【APUE】Chapter8 Process Control

    这章的内容比较多.按照小节序号来组织笔记的结构:再结合函数的示例带代码标注出来需要注意的地方. 下面的内容只是个人看书时思考内容的总结,并不能代替看书(毕竟APUE是一本大多数人公认的UNIX圣经). ...

  3. Cannot create a secure XMLInputFactory --CXF调用出错

    在调用方法前加上下面三句即可调用成功: Properties props = System.getProperties(); props.setProperty("org.apache.cx ...

  4. CSU-2110 Keeping Cool

    题目链接 http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2110 题目 Description Kevin has just got ...

  5. springbootday06 mysql

    一.MySql 1. 数据库概述 数据库( Database )是按照数据结构来组织.存储和管理数据的仓库 . 数据按照特定的格式存储起来,用户可以通过SQL (Structured Query La ...

  6. springboot04 Ajax json Jquery

    一.Ajax 1.同步&异步请求 在所有的请求响应交互世界里,我们有通常会划分出来两种形态的请求, 一种是同步请求.另一种是异步请求 .比如注册.登录.添加数据等等这些请求执行的就是同步请求, ...

  7. [译]如何禁止Requests库的log日志信息呢?

    原文来源: https://stackoverflow.com/questions/11029717/how-do-i-disable-log-messages-from-the-requests-l ...

  8. C++ 虚函数的内存分配

    1.无继承的普通类:   在有虚函数的情况下类会为其增加一个隐藏的成员,虚函数表指针,指向一个虚函数表,虚函数表里面就是类的各个虚函数的地址了.那么,虚函数表指针是以什么模型加入到类里面的,虚函数表里 ...

  9. poj 3436 网络流构图经典

    ACM Computer Factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6012   Accepted: 2 ...

  10. java中从实体类中取值会忽略的的问题

    在我们java Map中通过get来取值时会忽略的问题是:如果取得一个空值null时,那么.toString()时就会出错,而且不知道是什么原因. 现在我给的具体方法是用条件表达式先判断一下. 例: ...