Patrick and Shopping（距离比较）

                            Patrick and Shopping

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d
& %I64u

SubmitStatus

Description

Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is ad1
meter long road between his house and the first shop and ad2 meter long road between his house and the second shop. Also, there is a road of lengthd3
directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.



Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only
goal is to minimize the total distance traveled.

Input

The first line of the input contains three integersd1,
d2,
d3 (1 ≤ d1, d2, d3 ≤ 108) — the
lengths of the paths.

d1 is the length of the path connecting Patrick's house and the first shop;
d2 is the length of the path connecting Patrick's house and the second shop;
d3 is the length of the path connecting both shops.

Output

Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.

Sample Input

Input

10 20 30

Output

60

Input

1 1 5

Output

4

Hint

The first sample is shown on the picture in the problem statement. One of the optimal routes is: house


first shop

second shop


house.

In the second sample one of the optimal routes is: house


first shop

house


second shop

house

题意，给出两商店离家的距离，同时给出两商店间的距离，从家中出发，要经过两商店，再回到家中，求最短距离，中间可以回家。

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
int a,b,c,t;
while(~scanf("%d%d%d",&a,&b,&c))
{t=max(a,max(b,c));
if((a+b)<=t)
printf("%d\n",2*(a+b));
else if((a+c)<=t)
printf("%d\n",2*(a+c));
else if((b+c)<=t)
printf("%d\n",2*(b+c));
else printf("%d\n",a+b+c);
}
}