Python3 solution

lim = 3200
mark = bytearray(lim)
primes = [2]
for i in range(3, lim, 2):
    if mark[i]: continue
    primes.append(i)
    for j in range(3 * i, lim, 2 * i):
        mark[j] = 1

def factor(n):
    f = {}
    for p in primes:
        if p * p > n: break
        while n % p == 0:
            n //= p
            f[p] = f.get(p, 0) + 1
    if n > 1: f[n] = 1
    return f

T = int(input())
for t in range(T):
    n1, k1, n2, k2, n = map(int, input().split())
    if n1 == 0:
        f1 = 1 if k1 == 0 else 0
    else:
        f1 = pow(n1, k1, n) or n
    if n2 == 0:
        f2 = 1 if k2 == 0 else 0
    else:
        phi = 1
        for p, e in factor(n).items():
            phi *= (p - 1) * p ** (e - 1)
        f2 = pow(n2, k2, phi) or phi
    print(pow(f1, f2, n))

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static BigInteger solve(int n1, int k1, int n2, int k2, int n) { String s1 = String.valueOf(n1); String s2 = String.valueOf(n2); String s3 = String.valueOf(n); BigInteger b1, b2, b3; b1 = new BigInteger(s1); b2 = new BigInteger(s2); b3 = new BigInteger(s3); b1 = b1.pow(k1); b2 = b2.pow(k2); int i = b2.intValue(); BigInteger result = b1.pow(i); result = result.mod(b3); return result; }

Hints:

(1) Euler's theorem: if a and n are coprime positive integers, we have

a^phi(n) mod n = 1,

where phi() is Euler's totient function. Fermat's little theorem is a special case of this where n is prime.

(2) What if a and n are not coprime? Try this:

a mod n = (a/d mod n/d)*d,

where d=gcd(a,n) is the greatest common divisor of a and n.

These are the main ingredients of this problem. Watch out for boundary cases.

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
long long n1,k1,n2,k2,n,t,i,f,j,l;
cin>>t;
for(i=0;i<t;i++)
{cin>>n1>>k1>>n2>>k2>>n;
    if(n1==0 && k1==0)
        j=1;
    j=pow(n1,k1);
    if(n2==0 && k2==0)
        l=1;
    l=pow(n2,k2);
 if (j==0 && l==0)
     f=1;

    f=pow(j,l);
    cout<<f%n<<'\n';
}
giving error in some tests
dont know what is wrong with this code?