poj1213Fantasy of a Summation（找规律，优化代码）

最新推荐文章于 2021-07-09 10:54:52 发布

原创最新推荐文章于 2021-07-09 10:54:52 发布 · 581 阅读

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本文针对一个复杂的多重循环求和问题，提出了一种高效的算法优化方案。通过对原始代码的深入分析，发现其时间复杂度过高，进而引入数学方法简化计算过程，实现快速求解。

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Fantasy of a Summation

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

Submit Status

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

按代码写肯定会超时，那么该如何优化代码呢？

观察代码：

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;
  }
...
  }
  }
  }

代码有k重循环，简单模拟，设有3重循环，2个数，

1	0	0	0
2	0	0	1
3	0	1	0
4	0	1	1
5	1	0	0
6	1	0	1
7	1	1	0
8	1	1	1

共进行2^3次操做因为每个数的累加次数一样多的
它每个数的累加次数为(n^k)*k/n=(n^(k-1))*k,
则和就等于(sum(a[i])*(n^(k-1))*k)%mod

#include<stdio.h>
#include<algorithm>
#include<iostream>
#define  LL  long long
using namespace std;  
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}  
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}  
LL powmod(LL a,LL b,LL MOD)
{LL ans=1;
while(b)
{if(b%2)
ans=ans*a%MOD;
a=a*a%MOD;
b/=2;}
return ans;}   
int main()  
{  
    int t;  
    int cas=0;  
    scanf("%d",&t);  
    while(t--)  
    {  
        int n,k,mod;  
        scanf("%d%d%d",&n,&k,&mod);  
        LL ans=0;  
        for(int i=0;i<n;i++)  
        {  
            int a;  
            scanf("%d",&a);  
            ans=(ans+a)%mod;  
        }  
        LL num=powmod(n,k-1,mod);  
        printf("Case %d: ",++cas);  
        printf("%d\n",(ans*num*k)%mod);  
    }  
    return 0;  
}