题目链接:

A. Little Artem and Presents

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Artem got n stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.

How many times can Artem give presents to Masha?

Input
 

The only line of the input contains a single integer n (1 ≤ n ≤ 109) — number of stones Artem received on his birthday.

Output
 

Print the maximum possible number of times Artem can give presents to Masha.

Examples
 
input
1
output
1
input
2
output
1
input
3
output
2
input
4
output
3
Note

In the first sample, Artem can only give 1 stone to Masha.

In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.

In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.

In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.

题意:

问每次给个数,相邻的数不能一样,和为n,怎么才能使次数最多,最多为多少次;

思路

次数最多,那么就尽量为1和2了;

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+;
const ll inf=1e15;
const int N=1e5+; int main()
{ int n;
scanf("%d",&n);
if(n%==)printf("%d\n",n/*);
else if(n%==)printf("%d\n",n/*+);
else
{
printf("%d\n",n/*+);
}
return ;
}

codeforces 669A A. Little Artem and Presents(水题)的更多相关文章

  1. Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) A. Little Artem and Presents 水题

    A. Little Artem and Presents 题目连接: http://www.codeforces.com/contest/669/problem/A Description Littl ...

  2. codeforces 669B B. Little Artem and Grasshopper(水题)

    题目链接: B. Little Artem and Grasshopper time limit per test 2 seconds memory limit per test 256 megaby ...

  3. codeforces 669C C. Little Artem and Matrix(水题)

    题目链接: C. Little Artem and Matrix time limit per test 2 seconds memory limit per test 256 megabytes i ...

  4. codeforces Gym 100187L L. Ministry of Truth 水题

    L. Ministry of Truth Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/p ...

  5. Codeforces Round #185 (Div. 2) B. Archer 水题

    B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B D ...

  6. Educational Codeforces Round 14 A. Fashion in Berland 水题

    A. Fashion in Berland 题目连接: http://www.codeforces.com/contest/691/problem/A Description According to ...

  7. Codeforces Round #360 (Div. 2) A. Opponents 水题

    A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents ...

  8. Codeforces Round #190 (Div. 2) 水果俩水题

    后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟 ...

  9. Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)解题报告

    对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespac ...

随机推荐

  1. Go语言_RPC_Go语言的RPC

    一 标准库的RPC RPC(Remote Procedure Call,远程过程调用)是一种通过网络从远程计算机程序上请求服务,而不需要了解底层网络细节的应用程序通信协议.简单的说就是要像调用本地函数 ...

  2. win7 更改同步时间的网址

    Windows Registry Editor Version 5.00 [HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\D ...

  3. Examples osgparticleshader例子学习

    Examples osgparticleshader  粒子与shader的使用 参考文档 http://blog.csdn.net/csxiaoshui/article/details/234345 ...

  4. Opencv 图片边缘检测和最小外接矩形

    #include "core/core.hpp" #include "highgui/highgui.hpp" #include "imgproc/i ...

  5. Pearson product-moment correlation coefficient in java(java的简单相关系数算法)

    一.什么是Pearson product-moment correlation coefficient(简单相关系数)? 相关表和相关图可反映两个变量之间的相互关系及其相关方向,但无法确切地表明两个变 ...

  6. web微信开发

    群里接收消息时,使用广播,但需要刷新页面才能接收到广播内容. - 轮询: 定时每秒刷新一次,当群不活跃时,群里的每个客户端都在刷新,对服务端压力太大. - 长轮询:客户端连服务端,服务端一直不断开,也 ...

  7. HDU-3681-Prison Break(BFS+状压DP+二分)

    Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But one da ...

  8. javascript 高级编程系列 - 函数

    一.函数创建 1. 函数声明 (出现在全局作用域,或局部作用域) function add (a, b) { return a + b; } function add(a, b) { return a ...

  9. caffe训练自己的图片进行分类预测--windows平台

    caffe训练自己的图片进行分类预测 标签: caffe预测 2017-03-08 21:17 273人阅读 评论(0) 收藏 举报  分类: caffe之旅(4)  版权声明:本文为博主原创文章,未 ...

  10. windows下gVim 中文显示为乱码

    打开vimrc文件,在vim的安装目录下可以找到该文件: 或在windows下是在vim/gvim下输入:edit $vim/_vimrc. 在文件的末尾添加一句 "set fileenco ...