Hello coders, today we are going to solve **Travel Pass CodeChef Solution** whose Problem Code is **TRAVELPS**.

**Task**

Chef is going on a road trip and needs to apply for inter-district and inter-state travel e-passes. It takes ** A** minutes to fill each inter-district e-pass application and

**minutes for each inter-state e-pass application.**

*B*His journey is given to you as a binary string ** S** of length

**where**

*N***0**denotes crossing from one district to another district (which needs an inter-district e-pass), and a

**1**denotes crossing from one state to another (which needs an inter-state e-pass).

Find the total time Chef has to spend on filling the various forms.

**Input Format**

- The first line of the input contains a single integer
denoting the number of test cases. The description of*T*test cases follows.*T* - Each test case contains two lines of input.
- First line contains three space separated integers
,*N*and*A*.*B* - Second line contains the string
.*S*

**Output Format**

For each testcase, output in a single line the total time Chef has to spend on filling the various forms for his journey.

**Constraints**

**1 â‰¤ T â‰¤ 10**^{2}**1 â‰¤ N, A, B â‰¤ 10**^{2}**S**_{i}âˆˆ {â€² 0 â€²,â€² 1 â€²}

**Subtasks**

**Subtask #1 (100 points):** original constraints

**Sample Input 1**

```
3
2 1 2
00
2 1 1
01
4 2 1
1101
```

**Sample Output **1

** 2
2
5 **

**Explanation**

**Test case 1:** Chef needs total **2** inter-district e-passes, and he will be filling them in total **1 â‹… 2 = 2** minutes.

**Test case 3:** Chef needs total **1** inter-district e-pass and **3** inter-state e-passes, and he will be filling them in total **2 â‹… 1 + 1 â‹… 3 = 5** minutes.

**Solution – Travel Pass**

**C++**

#include <iostream> using namespace std; int main() { // your code goes here int t;cin>>t; while(t--){ int n,a,b; cin>>n>>a>>b; string s; cin>>s; int count0=0,count1=0; for(int i=0;i<n;i++){ if(s[i]=='0')count0++; else count1++; } cout << (a*count0 + b*count1) <<"\n"; } return 0; }

**Python**

# cook your dish here T = int(input()) for i in range(T): n, a, b = map(int, input().split()) string = input() count_1 = string.count('1') count_0 = string.count('0') total_cost = count_1 * b + count_0 * a print(total_cost)

**Java**

/* 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 sc = new Scanner(System.in); int t = sc.nextInt(); while(t-- > 0){ int n,a,b; n = sc.nextInt(); a = sc.nextInt(); b = sc.nextInt(); sc.nextLine(); String s = sc.nextLine(); char[] charArr = s.toCharArray(); int countZero=0,countOne=0; for(int i=0;i<n;i++){ if(charArr[i]=='0')countZero++; else countOne++; } System.out.println(a*countZero + b*countOne); } } }

**Disclaimer:** The above Problem **(Travel Pass)** is generated by **CodeChef **but the Solution is Provided by **CodingBroz**. This tutorial is only for** Educational** and **Learning** Purpose.