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You are given that a mango weighs X kilograms and a truck weighs Y kilograms. You want to cross a bridge that can withstand a weight of Z kilograms.
Find the maximum number of mangoes you can load in the truck so that you can cross the bridge safely.
For each test case, output in a single line the maximum number of mangoes that you can load in the truck.
Input: 4
2 5 11
4 10 20
1 1 1
6 40 90
Output: 3
2
0
8
Test case 1: You can load 3 mangoes at maximum. The total weight is 3\times 3×2+5=11≤11. Thus, the truck can safely cross the bridge with 3 mangoes. If you load 4 mangoes, the total weight is 4×2+5=13>11.
Test case 2: You can load 2 mangoes at maximum. The total weight is 2×4+10=18≤20. Thus, the truck can safely cross the bridge with 2 mangoes.
Test case 3: You can load 0 mangoes at maximum. The total weight is 0×1+1=1≤1. Thus, the truck can safely cross the bridge only if there are 0 mangoes.
Test case 4: You can load 8 mangoes at maximum. The total weight is 6×8+40=88≤90. Thus, the truck can safely cross the bridge with 8 mangoes.
import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner scn= new Scanner(System.in); //System.in is a standard input stream
int T= scn.nextInt();
for(int i=0; i<T; i++){
int X= scn.nextInt();
int Y= scn.nextInt();
int Z= scn.nextInt();
System.out.println((Z-Y)/X);
}
}
}
T = int(input())
for i in range(T):
X,Y,Z=map(int,input().split())
A = (Z-Y)//X
print(A)
#include<bits/stdc++.h>
using namespace std;
#define int long long int
int32_t main()
{
int t;
cin>>t;
while(t--)
{
int x,y,z;
cin>>x>>y>>z;
z-=y;
if(z<0)
z=0;
z/=x;
cout<<z<<endl;
}
}
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