LightOJ 1213 Fantasy of a Summation(快速幂)

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;
}
...
}
}
}

ans = sum * n^(k -1) * k

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#include <set>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 1010;

int T, n, total, cases = 0, K;
ll sum, mod, data[MAX_N];

ll quick_pow(ll a, ll b)
{
ll res = 1, tmp = a;
while(b){
if(b & 1) res = res * tmp % mod;
tmp = tmp * tmp % mod;
b >>= 1;
}
return res;
}

int main()
{
scanf("%d", &T);
while(T--){
sum = 0;
scanf("%d%d%lld", &n, &K, &mod);
for(int i = 0; i < n; i++){
scanf("%lld", &data[i]);
sum += data[i];
sum %= mod;
}
ll ans = quick_pow(n, K - 1) * K % mod * sum % mod;
printf("Case %d: %lld\n", ++cases, ans);
}
return 0;
}