Fantasy of a Summation LightOJ - 1213(找规律+快速幂)

最新推荐文章于 2021-05-20 08:47:03 发布

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本文介绍了一种名为Fantasy of a Summation的算法问题，该问题涉及多重循环中的求和运算，并通过优化方法实现高效的计算。具体地，文章探讨了如何通过分析循环特性简化计算过程，使用快速幂算法来解决大规模数据的处理问题。

Fantasy of a Summation

LightOJ - 1213

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 ... A_n-1, you have to write K nested loops and calculate the summation of all A_i 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 ... A_n-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

3 1 35000

1 2 3

2 3 35000

1 2

Sample Output

Case 1: 6

Case 2: 36

我们观察这个代码中的k次for循环可以知道语句 res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;
被执行了n^k次，而每次加k个数，所以总共下来一共会加k * n^k个数
同时我们还可以发现n个数中每一个数字被加的次数都是相同的，所以既然总共加了k * n^k个数，那么每个数字就加了
（k * n^k ） / n = k * n^(k-1) 次
所以总和 = a1 * k * n^(k-1) + a2 * k * n^(k-1) + …… +an * k * n^(k-1) =(a1+a2+……+an) * k * n^(k-1)
= sum * k * n^(k-1)
然后用快速幂求解即可
注意每一步都需要取模，求和sum的时候也要取模

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
ll q_pow(ll a,ll b,ll mod){
    ll ans = 1;
    while(b){
        if(b & 1)
            ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans % mod;
}
int main(){
    int t,cas = 0;
    scanf("%d",&t);
    while(t--){
        ll n,k,mod;
        ll sum = 0;
        cin >> n >> k >> mod;
        for(ll i = 1; i <= n; i++){
            ll x;
            cin >> x;
            sum = (sum + x) % mod;
        }
        ll ans = sum * k % mod;
        ans = ans * q_pow(n,k-1,mod) % mod;
        printf("Case %d: %d\n",++cas,ans);
    }
    return 0;
}