「CF478C」Table Decorations

2014年10月17日3,7610

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 test(s)

input

5 4 3

5 4 3

output

4

input

1 1 1

1 1 1

output

1

input

2 3 3

2 3 3

output

2

Note

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.

题解

此题乱搞的话很容易FST

赛后对着数据乱搞才过的。。。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
#define inf 1000000000
#define ll long long
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int a[5];
int ans;
int main()
{
    for(int i=1;i<=3;i++)a[i]=read();
    sort(a+1,a+4);

    int t=min(a[1],(a[3]-a[2])/2);
    a[1]-=t;a[3]-=t*2;ans+=t;

    ans+=a[1];a[2]-=a[1];a[3]-=a[1];

    t=min(a[2],a[3]-a[2]);
    a[2]-=t;a[3]-=t*2;ans+=t;
    if(a[2]>a[3])swap(a[2],a[3]);
    
    t=a[2]/3;ans+=t*2;
    a[2]-=t*3;a[3]-=t*3;
    if(a[2]*a[3]>=2)ans++;
    
    printf("%d\n",ans);
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<cstring>

#define inf 1000000000

#define ll long long

using namespace std;

int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int a[5];

int ans;

int main()

{

for(int i=1;i<=3;i++)a[i]=read();

sort(a+1,a+4);

int t=min(a[1],(a[3]-a[2])/2);

a[1]-=t;a[3]-=t*2;ans+=t;

ans+=a[1];a[2]-=a[1];a[3]-=a[1];

t=min(a[2],a[3]-a[2]);

a[2]-=t;a[3]-=t*2;ans+=t;

if(a[2]>a[3])swap(a[2],a[3]);

t=a[2]/3;ans+=t*2;

a[2]-=t*3;a[3]-=t*3;

if(a[2]*a[3]>=2)ans++;

printf("%d\n",ans);

return 0;

}

http://blog.csdn.net/caduca/article/details/40183233

当最大的气球数量的一半小于另外两种颜色的数量之和，可以组成全部颜色数量之和/3,否则全部组成为2+1，2为最多数量的颜色

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long i,j,t,n,m,l,r,k,z,y,x;
long long g,b,ans;
int main()
{
	 scanf("%I64d%I64d%I64d",&r,&g,&b);
	 ans=min(min(min((r+g+b)/3,r+g),r+b),b+g);
	 printf("%I64d\n",ans);
	 return 0;
}