2013 - ECJTU 暑期训练赛第七场-problem-F

F -O.O
Crawling in process...Crawling failedTime
Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatusPractice

CodeForces 90A

Description

A group of university students wants to get to the top of a mountain to have a picnic there. For that they decided to use a cableway.

A cableway is represented by some cablecars, hanged onto some cable stations by a cable. A cable is scrolled cyclically between the first and the last cable stations (the first of them is located
at the bottom of the mountain and the last one is located at the top). As the cable moves, the cablecar attached to it move as well.

The number of cablecars is divisible by three and they are painted three colors: red, green and blue, in such manner that after each red cablecar goes a green one, after each green cablecar goes a
blue one and after each blue cablecar goes a red one. Each cablecar can transport no more than two people, the cablecars arrive with the periodicity of one minute (i. e. every minute) and it takes exactly30 minutes for a cablecar
to get to the top.

All students are divided into three groups:r of them like to ascend only in the red cablecars,g of them prefer only the
green ones and b of them prefer only the blue ones. A student never gets on a cablecar painted a color that he doesn't like,

The first cablecar to arrive (at the moment of time0) is painted red. Determine the least time it will take all students to ascend to the mountain top.

Input

The first line contains three integersr,g andb (0 ≤ r, g, b ≤ 100).
It is guaranteed thatr + g + b > 0, it means that the group consists of at least one student.

Output

Print a single number — the minimal time the students need for the whole group to ascend to the top of the mountain.

Sample Input

Input

1 3 2

Output

34

Input

3 2 1

Output

33

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<cstdlib>
using namespace std;
int s[3];
int main()
{
int r,g,b,i,Time;
s[0]=1;
s[1]=2;
s[2]=0;
while(cin>>r>>g>>b)
{
Time=0;
i=0;
for(i=0;(r+g+b)>0;i=s[i])
{
if(i==0)
{
if(r>=2)
{
r-=2;
Time+=1;
}
else if(r>0)
{
r=0;
Time+=1;
}
else if(r==0)
{
Time+=1;
}
}
if(i==1)
{
if(g>=2)
{
g-=2;
Time+=1;
}
else if(g>0)
{
g=0;
Time+=1;
}
else if(g==0)
Time+=1;
}
if(i==2)
{
if(b>=2)
{
b-=2;
Time+=1;
}
else if(b>0)
{
b=0;
Time+=1;
}
else if(b==0)
{
Time+=1;
}
}
}
cout<<Time+30-1<<endl;
}
return 0;
}