Codeforces Round #441 (Div. 2)A. Trip For Meal

http://codeforces.com/problemset/problem/876/A

Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit’s and Owl’s houses is a meters, between Rabbit’s and Eeyore’s house is b meters, between Owl’s and Eeyore’s house is c meters.

For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit’s house and has a meal for the first time. Each time when in the friend’s house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend’s house, the supply of honey in other friend’s houses recover (most probably, they go to the supply store).

Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.

Input

First line contains an integer n (1 ≤ n ≤ 100) — number of visits.

Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit’s and Owl’s houses.

Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit’s and Eeyore’s houses.

Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl’s and Eeyore’s houses.

Output

Output one number — minimum distance in meters Winnie must go through to have a meal n times.

Examples

input

3
2
3
1

output

3

input

1
2
3
5

output

0

Note

In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit’s house, then in Owl’s house, then in Eeyore’s house. Thus he will pass the distance 2 + 1 = 3.

In the second test case Winnie has a meal in Rabbit’s house and that is for him. So he doesn’t have to walk anywhere at all.

题意

Winnie需要吃饭，一天吃ｎ顿，每顿饭在不同的地方吃
Winnie一开始在Rabbit家，Rabbit家距离Owl家ａ，距离Eeyore家ｂ；Owl家距离Eeyore家ｃ。问Winnie吃ｎ顿饭需最少走多远。

题解

一开始在Rabbit家，所以要想到Eeyore家必须得先到另外两家
在三家来回吃肯定不是最优解（除非距离都一样）
最优的是在距离最近的两家来回倒换
第一种，a*(n-1)
第二种，b*(n-1)
第三种，a+c*(n-2)或者b+c*(n-2)
我们比较这三者大小来确定最优解

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <cmath>using namespace std;#define inf 0x3f3f3f3fint main(){    int n;    int a,b,c,ans1,ans2=inf;    scanf("%d%d%d%d",&n,&a,&b,&c);    ans1=min(a*(n-1),b*(n-1));    if(n>=2)       ans2=min(a,b)+c*(n-2);    cout << min(ans1,ans2) << endl;}