CodeForces 478C Table Decorations
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values r, g and b will find the maximum number t of tables, that can be decorated in the required manner.
Input
The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.
Output
Print a single integer t — the maximum number of tables that can be decorated in the required manner.
Sample Input
5 4 3
4
1 1 1
1
2 3 3
2
Hint
In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
long long r,g,b;
long long a[];
long long i,j,k;
while(scanf("%I64d %I64d %I64d",&r,&g,&b)!=EOF)
{
long long s=;
a[]=r,a[]=g,a[]=b;
sort(a+,a+);
s=s+a[];
long long q=(a[]+a[])-a[]*,o;
o=q/;
if(o<=a[])
s=s+o;
else
s=s+a[];
printf("%I64d\n",s);
}
return ;
}
CodeForces 478C Table Decorations的更多相关文章
- Codeforces Round #273 (Div. 2)-C. Table Decorations
http://codeforces.com/contest/478/problem/C C. Table Decorations time limit per test 1 second memory ...
- 贪心 Codeforces Round #273 (Div. 2) C. Table Decorations
题目传送门 /* 贪心:排序后,当a[3] > 2 * (a[1] + a[2]), 可以最多的2个,其他的都是1个,ggr,ggb, ggr... ans = a[1] + a[2]; 或先2 ...
- C. Table Decorations(Codeforces Round 273)
C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...
- 【CODEFORCES】 C. Table Decorations
C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...
- codeforces 的 Codeforces Round #273 (Div. 2) --C Table Decorations
C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces Round #273 (Div. 2)C. Table Decorations 数学
C. Table Decorations You have r red, g green and b blue balloons. To decorate a single table for t ...
- cf478C Table Decorations
C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard ...
- CF-478C Table Decorations (贪心)
Table Decorations Time limit per test: 1 second Memory limit per test: 256 megabytes Problem Descrip ...
- 【Codeforces 478C】Table Decorations
[链接] 我是链接,点我呀:) [题意] 给你r,g,b三种颜色的气球 每张桌子要放3个气球 但是3个气球的颜色不能全都一样 (允许两个一样,或者全都不一样) 问你最多能装饰多少张桌子 [题解] 先把 ...
随机推荐
- 关闭微软对win10的推送
找到windows中安装的更新,卸载 KB3035583 这个更新即可
- 转:MyEclipse8.6插件安装方法
通常,我们可以用update来直接安装.但是myeclipse限制了中国区的下载和更新.所以我们只能用插件配置的方法来实现. MyEclipse8.6插件安装同Eclipse插件安装方式大致相同,如下 ...
- 【NOIP模拟赛】lover——心上人
心上人 [问题描述] 人到初三,总会遇到情感问题,比方说小 J 就喜欢上了小 W.于是小 J 就需要说一长串的话讨小 W 欢心.现在已知小 W 听到一些词就会很高兴,而且白听不厌,但她又讨厌小 J ...
- linux下串口控制
/* 本程序符合GPL条约 * Beneboy 2003-5-16 */ #include <stdio.h> // printf #include < ...
- linux设备驱动归纳总结(三):5.阻塞型IO实现【转】
本文转载自:http://blog.chinaunix.net/uid-25014876-id-60025.html linux设备驱动归纳总结(三):5.阻塞型IO实现 xxxxxxxxxxxxxx ...
- 不错的linux下通用的java程序启动脚本(转载)
转自:http://www.cnblogs.com/langtianya/p/4164151.html 虽然写起动shell的频率非常不高...但是每次要写都要对付一大堆的jar文件路径,新加jar包 ...
- Linux内核调试方法总结【转】
转自:http://my.oschina.net/fgq611/blog/113249 内核开发比用户空间开发更难的一个因素就是内核调试艰难.内核错误往往会导致系统宕机,很难保留出错时的现场.调试内核 ...
- Android剪切板传递数据传递序列化对象数据
一.剪切板的使用介绍 1. 剪切板对象的创建 使用剪切板会用到,ClipboardManager对象,这个对像的创建不可以使用构造方法,主要是由于没有提供public的构造函数(单例模式),需要使用A ...
- http://www.cnblogs.com/xqin/p/4862849.html
一.前言 半年前左右折腾了一个前后端分离的架子,这几天才想起来翻出来分享给大家.关于前后端分离这个话题大家也谈了很久了,希望我这个实践能对大家有点点帮助,演示和源码都贴在后面. 二.技术架构 这两年a ...
- asp.net 查询,导出
using System;using System.Configuration;using System.Data;using System.Linq;using System.Web;using S ...