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

巴特西

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

Problem:

Input

Output

Sample Input

Sample Output

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