A. Left-handers, Right-handers and Ambidexters
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are at a water bowling training. There are l people who play with their left hand, r people, who play with their right hand, and a ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input
The only line contains three integers l, r and a (0 ≤ l, r, a ≤ 100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Examples
inputCopy
1 4 2
outputCopy
6
inputCopy
5 5 5
outputCopy
14
inputCopy
0 2 0
outputCopy
0
Note
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can’t be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
int sum=0;
if(a==b)
{
sum=2*a+(c/2)*2;
}
else if(a<b)
{
if(a+c<b)
{
sum=2*(a+c);
}
else sum=2*b+(c-b+a)/2*2;
}
else
{
if(b+c<a)
{
sum=2*(b+c);
}
else sum=2*a+(c-a+b)/2*2;
}
printf("%d\n",sum);
return 0;
}
先分析a和b是否相等，如果相等，那就是2倍的再加上c/2*2；不然就单独分析，见代码