Question

这个问题出现在练习赛中：

计算 N 三角形数字，它也是一个平方数，模数为10006699. （1≤N≤10^ 18）最多有10 ^ 5个测试例。

我发现我可以用递归关系 T _i = 6T _i-1 - T _i-2 + 2 ，其中 T ₀ = 0 且 T ₁ = 1 。

我使用矩阵求幂来表示每个测试用例的大约O（log N）性能，但它显然太慢了，因为有10 ^ 5个测试用例。事实上，即使约束只是（1≤N≤10^ 6），这段代码太慢了，我只能进行O（N）预处理和O（1）查询。 / p>

我应该改变我对问题的处理方法，还是应该优化代码的某些部分？

#include <ios>
#include <iostream>
#include <vector>
#define MOD 10006699

/*
Transformation Matrix:

 0 1 0   t[i]     t[i+1]
-1 6 1 * t[i+1] = t[i+2]
 0 0 1     2        2
*/

std::vector<std::vector<long long int> > multi(std::vector<std::vector<long long int> > a, std::vector<std::vector<long long int> > b)
{
    std::vector<std::vector<long long int> > c(3, std::vector<long long int>(3));
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            for (int k = 0; k < 3; k++)
            {
                c[i][j] += (a[i][k] * b[k][j]) % MOD;
                c[i][j] %= MOD;
            }
        }
    }
    return c;
}

std::vector<std::vector<long long int> > power(std::vector<std::vector<long long int> > vec, long long int p)
{
    if (p == 1) return vec;
    else if (p % 2 == 1) return multi(vec, power(vec, p-1));
    else
    {
        std::vector<std::vector<long long int> > x = power(vec, p/2);
        return multi(x, x);
    }
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    long long int n;
    while (std::cin >> n)
    {
        if (n == 0) break;
        else
        {
            std::vector<std::vector<long long int> > trans;
            long long int ans;
            trans.resize(3);

            trans[0].push_back(0);  
            trans[0].push_back(1);
            trans[0].push_back(0);
            trans[1].push_back(-1);
            trans[1].push_back(6);
            trans[1].push_back(1);
            trans[2].push_back(0);
            trans[2].push_back(0);
            trans[2].push_back(1);

            trans = power(trans, n);

            ans = (trans[0][1]%MOD + (2*trans[0][2])%MOD)%MOD;

            if (ans < 0) ans += MOD;

            std::cout << ans << std::endl;
        }
    }
}

Answer 1

注意：我删除了旧答案，这更有用

对于问题，您似乎不可能为O（log N）创建更好的渐近算法。但是，可以对当前代码进行修改，这不会改善渐近时间但会提高性能

以下是对代码的修改，它产生相同的答案：

#include <ctime>
#include <ios>
#include <iostream>
#include <vector>
#define MOD 10006699

void power(std::vector<std::vector<long long int> >& vec, long long int p)
{
    if (p == 1)
        return;

    else if (p & 1)
    {
        std::vector<std::vector<long long int> > copy1 = vec;
        power(copy1, p-1);

        std::vector<std::vector<long long int> > copy2(3, std::vector<long long int>(3));
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
            {
                for (int k = 0; k < 3; k++)
                    copy2[i][j] += (vec[i][k] * copy1[k][j]) % MOD;
                copy2[i][j] %= MOD;
            }
        vec = copy2;

        return;
    }

    else
    {
        power(vec, p/2);

        std::vector<std::vector<long long int> > copy(3, std::vector<long long int>(3));
        for (int i = 0; i < 3; i++)
            for (int j = 0; j < 3; j++)
            {
                for (int k = 0; k < 3; k++)
                    copy[i][j] += (vec[i][k] * vec[k][j]) % MOD;
                copy[i][j] %= MOD;
            }
        vec = copy;

        return;
    }
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    long long int n;
    while (std::cin >> n)
    {
        std::clock_t start = std::clock();
        if (n == 0) break;

        std::vector<std::vector<long long int> > trans;
        long long int ans;
        trans.resize(3);

        trans[0].push_back(0);  
        trans[0].push_back(1);
        trans[0].push_back(0);
        trans[1].push_back(-1);
        trans[1].push_back(6);
        trans[1].push_back(1);
        trans[2].push_back(0);
        trans[2].push_back(0);
        trans[2].push_back(1);

        power(trans, n);

        ans = (trans[0][1]%MOD + (2*trans[0][2])%MOD)%MOD;
        if (ans < 0) ans += MOD;
        std::cout << "Answer: " << ans << std::endl;

        std::cout << "Time: " << (std::clock() - start) / (double)(CLOCKS_PER_SEC / 1000) << " ms" << std::endl;
    }
}

差异主要是：

c[i][j] %= MOD;的代码动作超出k循环
通过引用传递向量
较少的函数调用

如果我在main的while循环中放置相同的计时代码，就像我在代码中一样，将文件命名为＃34; before.cpp＆＃34;，将我的文件命名为＆＃34;之后.cpp＆＃34;，并连续运行10次并进行全面优化，然后这些是我的结果：

Alexanders-MBP:Desktop alexandersimes$ g++ before.cpp -O3 -o before
Alexanders-MBP:Desktop alexandersimes$ ./before 
1000000000000000000
Answer: 6635296
Time: 0.708 ms
1000000000000000000
Answer: 6635296
Time: 0.542 ms
1000000000000000000
Answer: 6635296
Time: 0.688 ms
1000000000000000000
Answer: 6635296
Time: 0.634 ms
1000000000000000000
Answer: 6635296
Time: 0.626 ms
1000000000000000000
Answer: 6635296
Time: 0.629 ms
1000000000000000000
Answer: 6635296
Time: 0.629 ms
1000000000000000000
Answer: 6635296
Time: 0.629 ms
1000000000000000000
Answer: 6635296
Time: 0.632 ms
1000000000000000000
Answer: 6635296
Time: 0.695 ms

Alexanders-MBP:Desktop alexandersimes$ g++ after.cpp -O3 -o after
Alexanders-MBP:Desktop alexandersimes$ ./after 
1000000000000000000
Answer: 6635296
Time: 0.283 ms
1000000000000000000
Answer: 6635296
Time: 0.287 ms
1000000000000000000
Answer: 6635296
Time: 0.27 ms
1000000000000000000
Answer: 6635296
Time: 0.27 ms
1000000000000000000
Answer: 6635296
Time: 0.266 ms
1000000000000000000
Answer: 6635296
Time: 0.265 ms
1000000000000000000
Answer: 6635296
Time: 0.266 ms
1000000000000000000
Answer: 6635296
Time: 0.267 ms
1000000000000000000
Answer: 6635296
Time: 0.21 ms
1000000000000000000
Answer: 6635296
Time: 0.208 ms

计算第N个三角形数字，也是方形数字

1 个答案: