1. Loop over A spoilage as(i) in [1;a(i)[ range;
2. Use GCD to reduce fractions;
3. Overall timing for n=5 should be near 0.01 s.

For a good 5 minutes there I thought i was going loopy haha... what the hell is this question.

include

include

include

include

include

include

using namespace std;

void Luxury_Hampers(int n, vector a, vector b){ long sum1=0,sum2=0; int sum_of_a_till_now=0; int sum_of_b_till_now=0;

int till_now_bad_a=0;
int till_now_bad_b=0;

int flag=0;
vector <int> Na;
vector <int> Nb;
for (int i=0; i<n; i++){
    float r1,r2;
    flag=0;
    sum_of_a_till_now+=a[i];
    sum_of_b_till_now+=b[i];
    for (int j=a[i]-1; j>0; j--){
        r1=(float)j/a[i];
        for (int k=1; k<=b[i]; k++){
            r2=(float)k/b[i];
            if (r2>r1){
                till_now_bad_a+=j;
                till_now_bad_b+=k;
                if (i>0){
                    float prev_a = (float)till_now_bad_a/sum_of_a_till_now;
                    float prev_b = (float)till_now_bad_b/sum_of_b_till_now;
                    if (prev_a>prev_b){
                        Na.push_back(j);
                        Nb.push_back(k);
                        flag=1;
                        break;
                    }
                    else{
                        till_now_bad_a-=j;
                        till_now_bad_b-=k;
                    }
                }
                else{
                    Na.push_back(j);
                    Nb.push_back(k);
                    flag=1;
                    break;
                }
            }
        }
        if (flag==1)
        break;
    }
}
for (int i=0; i<n; i++){
    sum1+=a[i];
    sum2+=b[i];
}
long suma=0,sumb=0;
for (int i=0; i<Na.size(); i++){
    suma+=Na[i];
    sumb+=Nb[i];
}

long num=suma*sum2;
long den=sumb*sum1;

if (num>den){
    int a=num%den;
    if (num%a==0&&den%a==0){
        num/=a;
        den/=a;
    }
    cout<<num<<"/"<<den;
}
else if (den>num){
    int a=den%num;
    if (den%a==0&&num%a==0){
        num/=a;
        den/=a;
    }
    cout<<num<<"/"<<den;
}  

}

int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int n; cin>>n; vector a; vector b; for (int i=0; i>x; a.push_back(x); } for (int i=0; i>x; b.push_back(x); }

Luxury_Hampers(n,a,b);
return 0;

}

passed 6/16 test cases. failed 10/16 due to timeout. finding a way to optimize it. any suggestion?

from itertools import *
import math
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
n_a=[]
n_b=[]
sa=sum(a)
sb=sum(b)
flag=0
for i in range(n):
    n_a.append(list(range(1,a[i]+1)))   """make a list of all possible values of number of bad products for each product by A"""
    n_b.append(list(range(1,b[i]+1)))    #same for B
    
for i in product(*n_a):      """iterate through all the possible combinations of number of bad product by A"""
     for j in product(*n_b):  """iterate through all the possible combinations of number of bad product by B"""
           count=0                """counting the number of times ratio is maintained for the current combination number of bad products by A and B """
           p=sum(i)*sb            """expression you will get by putting the given condition in a equation (A-overall/B-overall) spoilage ratio"""
           q=sum(j)*sa
           m=math.gcd(p,q)  """reduce the fraction to its lowest terms"""
           p,q=p//m,q//m          
 
           if (p/q)<=1:           """less than 1 then continue to next iteration"""
                  continue

           for k in range(n):     """each pair of kth product check the equality of ratio """
                  r=(j[k]*a[k])
                  s=(i[k]*b[k])         """expression you will get by putting the given condition in a equation (B-kth/A-kth) per product spoilage ratio"""
                  
                  t=math.gcd(r,s)      """reduce the fraction to its lowest terms"""
                  r,s=r//t,s//t

                   if r!=p or s!=q :     """not equal then break the loop and go for next combination of i and j (change j)"""
                             break
                   else:
                              count+=1          """equal then increase count"""
             
                   if count==n:          """the ratio is same for all the n products then print the ratio and exit all the loops"""
                               flag=1
                               print("{}/{}".format(r,s))
                               break

           if flag==1:
                 break

    if flag==1:
        break

problem doesn,t contain proper data as spoilage data is not provided