如何选择前n个项? [英] How to select the first n terms?

问题描述

for(i=0;i<500;i++)
{
    x=valuex[i];
    y=valuey[i];

    A[i][0]=x+y;
    A[i][1]=x*x+3*y;
    A[i][2]=x+2*y*x;
    ……
    A[i][99]=x/2+20*y;
}

在上面的代码中，有100个项:A [i] [0]〜A [i] [99]，实际上，我想使用A [i] [j]的前n个项，而"n"是用户输入的数字，我如何有效地实现这一点，谢谢您的帮助.
最好的祝愿.

In the above code,there are 100 terms:A[i][0]~A[i][99], In fact, I want to use the first n terms of A[i][j],and "n" is the input number by user, how can I realize this efficiently,Thank you for your help.
Best wishes.

推荐答案

很好的问题，尽管我看不到真正的应用程序是什么.当我正确理解您的问题后，您只想计算
的前n个项

Nice question, although I don''t see what the real application is. When I understood your question correctly, you want to compute only the first n terms of

A[i][0]=x+y;
A[i][1]=x*x+3*y;
A[i][2]=x+2*y*x;
……
A[i][99]=x/2+20*y;

因此n是j的极限.正确吗?

至少有两种主要方法可以实现此目标:

(1)插入if语句

thus n being the limit for j. Correct?

There are (at least) two principal approaches of how you can go about that:

(1) Interspersing if statements

while (1)
{
    if (0 >= n) break;
    A[i][0]=x+y;

    if (1 >= n) break;
    A[i][1]=x*x+3*y;

    if (2 >= n) break;
    A[i][2]=x+2*y*x;
     ……

    if (99 >= n) break;
    A[i][99]=x/2+20*y;
    break;
}

这不是很优雅，但可能比方法2更快:

(2)将算术项放在函数数组中，然后在该数组上循环.

That is not elegant, but probably faster than approach number 2:

(2) Placing the arithmetic terms in an array of functions and then looping over that array.

typedef double PolyFunc (double x, double y);

double f0 (double x, double y)
    {return x + y;}

double f1 (double x, double y)
    {return x*x + 3*y;}

...

static PolyFunc* sFunctions[100] = {&f0, &f1, ...};

void ComputeArray (int n)
{
    ...
    for (int j = 0; j < n; ++j)
        A[i][j] = sFunctions[j] (x, y);
}

希望对您有所帮助.

Hope that helps.

关于您的问题有很多问题...
但是...

There are many questions about your question...
but...

switch (n)
{
case 99: A[i][99]=x/2+20*y;
...
case 2:   A[i][2]=x+2*y*x;
case 1:   A[i][1]=x*x+3*y;
case 0:   A[i][0]=x+y;
}

会很快.使用switch让编译器对if变体进行一些优化.
僵尸程序可能会更快一些:

will be very fast. Using switch let the compilers make some optimization over the if variant.
A bit confusing bot probably faster:

switch (99-n)
{
case 0:   A[i][99]=x/2+20*y;
...
case 97:   A[i][2]=x+2*y*x;
case 98:   A[i][1]=x*x+3*y;
case 99:   A[i][0]=x+y;

}

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