循环所有数字组合 [英] looping all number combinations

问题描述

我的工作是应该全自动测试值，看看是否有一个意外的错误测试功能。

I'm working on a test function that is supposed to test values automaticly to see if there is an unexpected error.

有一个应该具有-999之间的值范围为1000六大持有人

There are six "holders" that is supposed to have a value range between -999 to 1000

这是我试了一下：

    #include <stdlib.h>
#include <stdio.h>

int inuti(double x, double y, double x1, double y1, double x2, double y2) {
    int x_inuti;
    int y_inuti;

    if (x1 < x2)
        x_inuti = x > x1 && x < x2;
    else
        x_inuti = x > x2 && x < x1;
    if (y1 < y2)
        y_inuti = y > y1 && y < y2;
    else
        y_inuti = y > y2 && y < y1;
    return x_inuti && y_inuti;
}

int main(void) {
    double x, y, x1, y1, x2, y2;
    int resultat;
    x = -999;
    y = 1000;
    x1 = -999;
    y1 = 1000;
    x2 = -999;
    y2 = 1000;

    while (x<1000 && y>-999 && x1<1000 && y2 >-999 && x2<1000 && y2>-999) {
    x++;
    y--;
    y1++;
    x1--;
    x2++;
    y2--;


            printf("point x-value: %.1f \n", x);

            printf("point y-value: %.1f \n", y);

            printf("\n");

            printf("corner side x-value: %.1f \n", x1);


            printf("corner side y-value: %.1f \n", y1);

            printf("\n");

            printf("other corner side x-value%.1f \n", x2);


            printf("other corner side y-value%.1f \n", y2);

            printf("\n");


        resultat = inuti(x, y, x1, y1, x2, y2);

        if (resultat == 1)
            printf("point was inside rectangle.\n");
        else
            printf("point was outside rectangle.\n");

        printf("\n");
        printf("\n");
    }
        getchar();
    return 0;
}

然而，它并不包括所有的组合，它甚至超过了最大和最小数目。

However it doesn't include all combinations and it even exceeds the maximum and minimum number.

有谁知道循环更好的方法所有组合？

Does anyone know a better way to loop all combinations?

我AP preciate所有帮助

I appreciate all help

感谢

推荐答案

您会需要在6个不同的循环测试所有可能的不同组合。

You're going to need at 6 different loops to test all different possible combinations.

for (x = -999; x <= 1000; x++) {
    for (y = -999; y < 1000; y++) {
        for (x1 = -999; x1 <= 1000; x1++) {
            // etc...
        }
    }
}

但要知道，有2000个 ⁶ = 64,000,000,000,000,000,000不同的组合，意为code将pretty太多写不完执行（如果每个测试需要1微秒，在$ C $按c完成后约2028080年执行）。更好的选择是尝试测试，你知道输出的具体意见。请确保您测试的边缘情况了。

However be aware that there are 2000⁶ = 64,000,000,000,000,000,000 different combinations, meaning the code will pretty much never finish executing (if each test takes 1 microsecond, the code will finish executing after about 2,028,080 years). A better option is to try testing specific inputs where you know the output. Make sure you test the edge cases too.

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