Light OJ 1213 Fantasy of a Summation （快速幂+规律）

本文链接：https://blog.youkuaiyun.com/yao1373446012/article/details/52088534

本文描述了一个关于在梦中见到的复杂代码的问题，该代码涉及计算特定数组在多个嵌套循环下的所有元素组合之和，并提供了两种解决方案。输入包括多组测试用例，每组包含整数n、K及MOD，随后是一系列整数。

E - Fantasy of a Summation

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1213

uDebug

Appoint description:

Description

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

思路：模拟一下找规律，你会发现 sum*n^(k-1)*k

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
using namespace std;
LL mod(LL a,LL b,LL c)
{
    LL res=1%c;
	LL t=a%c;
	while(b)
	{
		if(b%2)
		{
			res=res*t%c;
		}
		t=t*t%c;
		b=b/2;
	 } 
	 return res;
}
int main()
{
	LL n,k,t,m,i,j,a;
	scanf("%lld",&t);
	LL l=1;
	while(t--)
	{
		scanf("%lld%lld%lld",&n,&k,&m);
		LL sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%lld",&a);
			sum+=a;
			sum=sum%m;
		}
		LL ans=((sum*mod(n,k-1,m))%m)*k%m;
		printf("Case %lld: %lld\n",l++,ans);
	}
	return 0;
}