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Janmansh is at the fruit market to buy fruits for Chingari. There is an infinite supply of three different kinds of fruits with prices A, B and C.
He needs to buy a total of X fruits having at least 2 different kinds of fruits. What is the least amount of money he can spend to buy fruits?
For each test case, output the least amount of money he can spend to buy fruits.
Input:
2
2 1 1 1
3 4 3 2
Output:
2
7
Test case-1: He can buy any two fruits of different kinds for a total price of 2.
Test case-2: He can buy 1 fruit of price 3 and 2 fruits of price 2 for a total price of 7.
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;cin>>t;
while(t--)
{
int x,a,b,c;
cin>>x>>a>>b>>c;
int d[3]={a,b,c};
sort(d,d+3);
cout<<d[0]*(x-1)+d[1]<<endl;
}
return 0;
}
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner in=new Scanner(System.in);
int te=in.nextInt();
while(te-->0){
int n=in.nextInt();
int ar[]=new int[3];
for(int i=0;i<3;i++)
ar[i]=in.nextInt();
Arrays.sort(ar);
int res=(n-1)*ar[0];
res+=ar[1];
System.out.println(res);
}
}}
# cook your dish here
t=int(input())
for i in range(t):
x,a,b,c=map(int,input().split())
m=min((a+b),(b+c),(c+a))
m1=min(a,b,c)
print(m+(x-2)*m1)
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