Как переделать код под себя? язык программирования С

Уважаемые, форумчане, прошу огромной просьбой помочь. Поступил на 1 курс "ПИ" и тут сходу в первом же месяце такое сложное задание, описание четко не помню, но звучало так: реализовать функции вычисления площади по составной формуле Симпсона, не суть. Однокурсник написал код, но нужно его переделать, дабы не был похож, чтоб не спалиться, помогите.

#include <math.h>
#include <string.h>

#define PI  3.14159

#pragma warning(disable : 4996)
#pragma warning(disable : 6031)

char answer[6][25];
double a, b;

double methode_rectangle(int n) {
    double x, h, integral = 0;
    h = (b - a) / n;
    x = a;
    for (int i = 0; i < n; i++) {
        integral += sin(x + h / 2) * h;
        x += h;
    }
    return integral;
}

double methode_simps(int n) {
    double x, h, integral;
    h = (b - a) / 2 / n;
    integral = sin(a) + sin(b);
    x = a + h;
    for (int i = 1; i < 2 * n; i++) {
        if (i % 2 == 0) integral += 2 * sin(x);
        else integral += 4 * sin(x);
        x += h;
    }
    integral *= h / 3;
    return integral;
}

void vizov(int a[6]) {
    char intermediate_line[15];
    for (int i = 0; i < 6; i++) {
        int j = 0;
        double integral_rectangle, integral_simpson;
        integral_rectangle = methode_rectangle(a[i]);
        integral_simpson = methode_simps(a[i]);
        sprintf(intermediate_line, "%d", a[i]);
        strcat(answer[i], intermediate_line);
        strcat(answer[i], " ");
        sprintf(intermediate_line, "%2.5f", integral_rectangle);
        strcat(answer[i], intermediate_line);
        strcat(answer[i], " ");
        sprintf(intermediate_line, "%2.5f", integral_simpson);
        strcat(answer[i], intermediate_line);
    }
}

void entering_numbers() {
    printf("Enter interval's left border: ");
    while (!(scanf("%lf", &a))) {
        printf("Error! Enter again: ");
        while (getchar() != '\n');
    }
    printf("Enter interval's right border: ");
    while (!(scanf("%lf", &b))) {
        printf("Error! Enter again: ");
        while (getchar() != '\n');
    }
}

int main()
{
    int number_of_segments[] = {5, 10, 20, 100, 500, 1000};
    entering_numbers();
    vizov(number_of_segments);
    for (int i = 0; i < 6; i++) {
        printf("%s\n", answer[i]);
    }
    return 0;
}

Ответы (1 шт):

Автор решения: Fat-Zer

StackOverflow не является сервисом обфрускации

Если воспользоваться одним из ресурсов, которые являются сервисами обфускации, то получится нечто такое:

#include <stdio.h>
#include <math.h>
#include <string.h>
#define PI  3.14159
char [25] o_f029e0247ceac467341197e74f5ebc0b[(0x0000000000000032 + 0x0000000000000219 + 0x0000000000000819 - 0x0000000000000A4B)]=(0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12);double o_67e963479f526c0727815a005e394bbf;double o_0fbe522679ff424694b3034b41bf4e8f;double o_65e6bccdfac523737c2c07670a06bd95(int o_f4609075a04cd29001299e4450ccf856){double o_b99b66b10ff11bf276f79fef1d7bc4af,o_9acfc00fc1c3f0baa21d9c6e20ea74ec,o_5ec8578bfba6949be2124472b5035190=(0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);o_9acfc00fc1c3f0baa21d9c6e20ea74ec = (o_0fbe522679ff424694b3034b41bf4e8f - o_67e963479f526c0727815a005e394bbf) / o_f4609075a04cd29001299e4450ccf856;o_b99b66b10ff11bf276f79fef1d7bc4af = o_67e963479f526c0727815a005e394bbf;for (int o_2a154e440f9265ed0151789fe7e3d9c2=(0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);(o_2a154e440f9265ed0151789fe7e3d9c2 < o_f4609075a04cd29001299e4450ccf856) & !!(o_2a154e440f9265ed0151789fe7e3d9c2 < o_f4609075a04cd29001299e4450ccf856);o_2a154e440f9265ed0151789fe7e3d9c2++){o_5ec8578bfba6949be2124472b5035190 += sin(o_b99b66b10ff11bf276f79fef1d7bc4af + o_9acfc00fc1c3f0baa21d9c6e20ea74ec / (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06)) * o_9acfc00fc1c3f0baa21d9c6e20ea74ec;o_b99b66b10ff11bf276f79fef1d7bc4af += o_9acfc00fc1c3f0baa21d9c6e20ea74ec;};return o_5ec8578bfba6949be2124472b5035190;};double o_3111478fdda083a6ef0bd8c10611dbc7(int o_70039496f897e60ef2ec4863cf9f47dc){double o_40df2e9ec2cef140f3167f2e516bed1e,o_d33649045cba9c3bd0c5934c8c9362d3,o_f96bbc9df12ba25579596202ad58fe85;o_d33649045cba9c3bd0c5934c8c9362d3 = (o_0fbe522679ff424694b3034b41bf4e8f - o_67e963479f526c0727815a005e394bbf) / (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06) / o_70039496f897e60ef2ec4863cf9f47dc;o_f96bbc9df12ba25579596202ad58fe85 = sin(o_67e963479f526c0727815a005e394bbf) + sin(o_0fbe522679ff424694b3034b41bf4e8f);o_40df2e9ec2cef140f3167f2e516bed1e = o_67e963479f526c0727815a005e394bbf + o_d33649045cba9c3bd0c5934c8c9362d3;for (int o_0e9da0083206357b7e195b81c59926ad=(0x0000000000000002 + 0x0000000000000201 + 0x0000000000000801 - 0x0000000000000A03);(o_0e9da0083206357b7e195b81c59926ad < (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06) * o_70039496f897e60ef2ec4863cf9f47dc) & !!(o_0e9da0083206357b7e195b81c59926ad < (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06) * o_70039496f897e60ef2ec4863cf9f47dc);o_0e9da0083206357b7e195b81c59926ad++){if (!(o_0e9da0083206357b7e195b81c59926ad % (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06) ^ 0x0000000000000000))o_f96bbc9df12ba25579596202ad58fe85 += (0x0000000000000004 + 0x0000000000000202 + 0x0000000000000802 - 0x0000000000000A06) * sin(o_40df2e9ec2cef140f3167f2e516bed1e);else o_f96bbc9df12ba25579596202ad58fe85 += (0x0000000000000008 + 0x0000000000000204 + 0x0000000000000804 - 0x0000000000000A0C) * sin(o_40df2e9ec2cef140f3167f2e516bed1e);;o_40df2e9ec2cef140f3167f2e516bed1e += o_d33649045cba9c3bd0c5934c8c9362d3;};o_f96bbc9df12ba25579596202ad58fe85 *= o_d33649045cba9c3bd0c5934c8c9362d3 / (0x0000000000000006 + 0x0000000000000203 + 0x0000000000000803 - 0x0000000000000A09);return o_f96bbc9df12ba25579596202ad58fe85;};void o_1843b40a765a8c97f52337cb51c98823(int o_878fb4de640e861f052de3d45afeeaad[(0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12)]){char o_38b03891c238c93f73dfdf871d4a4442[(0x000000000000001E + 0x000000000000020F + 0x000000000000080F - 0x0000000000000A2D)];for (int o_4168f7906ca9d7fb0aca3a692fb8fd41=(0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);(o_4168f7906ca9d7fb0aca3a692fb8fd41 < (0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12)) & !!(o_4168f7906ca9d7fb0aca3a692fb8fd41 < (0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12));o_4168f7906ca9d7fb0aca3a692fb8fd41++){int o_a70707a2ffc70c24c4552253de2d767f=(0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);double o_2697621e27b5a19f5956b8fcb59d04b4,o_38a21b7958f16c6a5153eed82d23e50d;o_2697621e27b5a19f5956b8fcb59d04b4 = o_65e6bccdfac523737c2c07670a06bd95(o_878fb4de640e861f052de3d45afeeaad[o_4168f7906ca9d7fb0aca3a692fb8fd41]);o_38a21b7958f16c6a5153eed82d23e50d = o_3111478fdda083a6ef0bd8c10611dbc7(o_878fb4de640e861f052de3d45afeeaad[o_4168f7906ca9d7fb0aca3a692fb8fd41]);sprintf(o_38b03891c238c93f73dfdf871d4a4442,"\x25""d",o_878fb4de640e861f052de3d45afeeaad[o_4168f7906ca9d7fb0aca3a692fb8fd41]);strcat(o_f029e0247ceac467341197e74f5ebc0b[o_4168f7906ca9d7fb0aca3a692fb8fd41],o_38b03891c238c93f73dfdf871d4a4442);strcat(o_f029e0247ceac467341197e74f5ebc0b[o_4168f7906ca9d7fb0aca3a692fb8fd41],"\x20""");sprintf(o_38b03891c238c93f73dfdf871d4a4442,"\x25""2\0565\x66""",o_2697621e27b5a19f5956b8fcb59d04b4);strcat(o_f029e0247ceac467341197e74f5ebc0b[o_4168f7906ca9d7fb0aca3a692fb8fd41],o_38b03891c238c93f73dfdf871d4a4442);strcat(o_f029e0247ceac467341197e74f5ebc0b[o_4168f7906ca9d7fb0aca3a692fb8fd41],"\x20""");sprintf(o_38b03891c238c93f73dfdf871d4a4442,"\x25""2\0565\x66""",o_38a21b7958f16c6a5153eed82d23e50d);strcat(o_f029e0247ceac467341197e74f5ebc0b[o_4168f7906ca9d7fb0aca3a692fb8fd41],o_38b03891c238c93f73dfdf871d4a4442);};};void o_83fd9cae134fbde6c8658948341e6d70(){printf("\x45""n\164e\x72"" \151n\x74""e\162v\x61""l\047s\x20""l\145f\x74"" \142o\x72""d\145r\x3A"" ");while (!(scanf("\x25""l\146",&o_67e963479f526c0727815a005e394bbf))){printf("\x45""r\162o\x72""!\040E\x6E""t\145r\x20""a\147a\x69""n\072 ");while (getchar() != '\n');;};printf("\x45""n\164e\x72"" \151n\x74""e\162v\x61""l\047s\x20""r\151g\x68""t\040b\x6F""r\144e\x72"":\040");while (!(scanf("\x25""l\146",&o_0fbe522679ff424694b3034b41bf4e8f))){printf("\x45""r\162o\x72""!\040E\x6E""t\145r\x20""a\147a\x69""n\072 ");while (getchar() != '\n');;};};int main(){int o_ea8f388c12483cdd3835d9b2a1114240[]={5, 10, 20, 100, 500, 1000};o_83fd9cae134fbde6c8658948341e6d70();o_1843b40a765a8c97f52337cb51c98823(o_ea8f388c12483cdd3835d9b2a1114240);for (int o_3d48f3e14d80ac4b22b8bee6e1bc8947=(0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);(o_3d48f3e14d80ac4b22b8bee6e1bc8947 < (0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12)) & !!(o_3d48f3e14d80ac4b22b8bee6e1bc8947 < (0x000000000000000C + 0x0000000000000206 + 0x0000000000000806 - 0x0000000000000A12));o_3d48f3e14d80ac4b22b8bee6e1bc8947++){printf("\x25""s\012",o_f029e0247ceac467341197e74f5ebc0b[o_3d48f3e14d80ac4b22b8bee6e1bc8947]);};return (0x0000000000000000 + 0x0000000000000200 + 0x0000000000000800 - 0x0000000000000A00);};
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