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
- 2. 星际争霸之php面向对象(二)
题记==============================================================================本php设计模式专辑来源于博客(jymo ...
- html里那些细节
target="_top",项目是frameset形式写的,用这个在跳转的时候从父框架跳转
- Java简单数据类型转换
1. Integer<---String (1) Integer x = new Integer(Integer.parseInt(String)); 2. Integer<--- ...
- [转]jexus的安装
转自http://www.cnblogs.com/xiaodiejinghong/p/3720921.html 这是一个集成了 mono3.4.0 和 jexus5.6.0 的 jexus+mono ...
- MySQL 5.7 SYS系统SCHEMA
版权声明:本文为博主原创文章,未经博主允许不得转载. 在说明系统数据库之前,先来看下MySQL在数据字典方面的演变历史:MySQL4.1 提供了information_schema 数据字典.从此可以 ...
- php中method_exists()和is_callable()如何进行语句判断
method_exists()和is_callable()方法进行判断.那么两则区别是什么呢? 已知类文件如下: class Student{private $alias=null;private $ ...
- HDU:Integer Inquiry
#include"stdio.h" #include"stdlib.h" #include"string.h" #define N 105 ...
- [团队项目]SCRUM项目5.0
5.0--------------------------------------------------- 1.团队成员完成自己认领的任务. 2.燃尽图:理解.设计并画出本次Sprint的燃尽图的理 ...
- 刚体Collider包围测试
测试结果为会自动排出修正坐标(之前位于中心): 2016/2/29补充: 如果外面大的Cube相对小的Cube质量很高,会弹出且不出现移动(已锁住弹出物旋转,如果不锁会飞出去): 如果没有足够的空间排 ...
- Uva 10305 给任务排序
题目链接:https://uva.onlinejudge.org/external/103/10305.pdf 紫书P167 拓扑排序. dfs——从一个点出发,dfs 与之相连的所有点,把本身放入到 ...