Fantasy of a Summation

If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is the code I saw in my dream.

#include <stdio.h>
int cases, caseno;
int n, K, MOD;
int A[1001];

int main() {

    scanf("%d", &cases);

    while( cases-- ) {

        scanf("%d %d %d", &n, &K, &MOD);

        int i, i1, i2, i3, ... , iK;

        for( i = 0; i < n; i++ ) scanf("%d", &A[i]);

        int res = 0;

        for( i1 = 0; i1 < n; i1++ ) {

            for( i2 = 0; i2 < n; i2++ ) {

                for( i3 = 0; i3 < n; i3++ ) {

                    ...

                    for( iK = 0; iK < n; iK++ ) {

                        res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;

                    }

                    ...

                }

            }

        }

        printf("Case %d: %d\n", ++caseno, res);

    }

    return 0;
}

Actually the code was about: 'You are given three integers n,
K, MOD and n integers:
A0, A1, A2 ... An-1, you have to write
K nested loops and calculate the summation of all Ai where
i is the value of any nested loop variable.'

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with three integers: n (1 ≤ n ≤ 1000), K (1 ≤ K < 231), MOD (1 ≤ MOD ≤ 35000). The next line contains
n non-negative integers denoting A0, A1, A2 ... An-1.
Each of these integers will be fit into a 32 bit signed integer.

Output
For each case, print the case number and result of the code.

Sample Input
2

3 1 35000

1 2 3

2 3 35000

1 2

Sample Output
Case 1: 6

Case 2: 36

题意：给出一段代码看懂简化下，这题就是说有k层循环每层遍历n个数字，最后把k层遍历到的数字全部加和取模。
思路：
稍微推下公式就有res=n^k*k/n*sum%mod;就是一共有n^k种组合，每种组合k个数字，平均分给n个数字，之后快速幂解决。

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int MAXN = 1e7+10;
int a[1003];
LL qp(LL n,LL m,LL mod)
{
LL ans=1;
while(m)
{
if(m&1)
{
ans=(ans*n)%mod;
}
n=(n*n)%mod;
m=m>>1;
}
return ans;
}
int main()
{
int t;
cin>>t;
int g=0;
while(t--)
{
++g;
LL n,k,mod;
cin>>n>>k>>mod;
LL sum=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
sum+=a[i];
}
LL sum1=qp(n,k-1,mod),sum2;
// cout<<"**"<<sum<<endl;
// cout<<sum1<<endl;
sum2=sum1%mod*k%mod*sum%mod;
printf("Case %d: %lld\n",g,sum2);
}
return 0;
}