Fantasy of a Summation （LightOJ - 1213）（快速幂+简单思维）

题解：根据题目给的程序，就是计算给的这个序列，进行k次到n的循环，每个数需要加的次数是k*n^(k-1)，所以快速幂取模，算计一下就可以了。

#include <bits/stdc++.h>

using  namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f3f;
long long pow_mod(ll a, ll k, ll mod)
{
    ll ans = 1;
    while(k)
    {
        if(k%2)
            ans *= a;
        ans %= mod;
        a = a * a;
        a %= mod;
        k /=2;
    }
    return ans;
}
int main()
{
    int T;
    ll n,k,mod,x,sum;
    while(~scanf("%d",&T))
    {
        int cas = 1;
        while(T--)
        {
            sum = 0;
            scanf("%lld%lld%lld",&n,&k,&mod);
            for(ll i = 0; i < n; i ++)
            {
                scanf("%lld",&x);
                sum += (x * (k  * pow_mod(n,k-1,mod)%mod)%mod);
                sum %= mod;
            }
            printf("Case %d: %lld\n",cas++, sum);
        }
    }
    return 0;
}

Problem:

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