Полностью сделал задачу 6 на вычислительную геометрию

Session/4Ex/main.c

@@ -0,0 +1,93 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+
+#define eps 1.e-6
+
+int ctv(const void * unum1, const void * unum2);
+double * getArray(FILE * file);
+int isContains(double * arr1, int len1, double * arr2, int len2);
+int write(const char * str);
+
+int ctv(const void * unum1, const void * unum2) {
+	double num1 = *(double *)unum1;	
+	double num2 = *(double *)unum2;
+	
+	if (num1 - num2 > eps) return 1;
+	else if (num2 - num1 > eps) return -1;
+	else return 0;
+}
+
+double * getArray(FILE * file) {
+	int size = 2, i = 0;
+	double * arr = (double *)malloc(size * sizeof(double));
+	double current;
+
+	if (arr == NULL) return NULL;
+
+
+	while (fscanf(file, "%lf", &current) == 1) {
+		if (size == (i + 1)) {
+			arr = (double *)realloc(arr, (size *= 2) * sizeof(double));
+			if (arr == NULL) return NULL;
+		}
+		arr[++i] = current;
+	}
+
+	arr[0] = i;
+	return arr;
+}
+
+int isContains(double * arr1, int len1, double * arr2, int len2) {
+	for (int i = 0, k = 0; i < len2; ++i, ++k) {
+		if (fabs(arr2[i] - arr1[k]) < eps) {
+			if (k + 1 == len1) return 1;
+		} else if (len2 - i <= len1) return 0;
+		else k = -1;
+	}
+	
+	return 0;
+}
+
+int write(const char * str) {
+	FILE * file = fopen("output.txt", "w");
+	
+	if (file == NULL) return -1;
+	if (fprintf(file, "%s", str) < 1) {fclose(file); return -1;}
+	if (fclose(file) != 0) return -1;
+
+	return 0;
+}
+
+int main(void) {
+	FILE * file = fopen("ina.txt", "r");
+	double *arr1, *arr2;
+	
+	if (file == NULL) return -1;
+	arr1 = getArray(file);
+	if (arr1 == NULL) {fclose(file); return -1;}
+	if (fclose(file) != 0) return -1;
+
+	file = fopen("inb.txt", "r");
+	if (file == NULL) return -1;
+	arr2 = getArray(file);
+	if (arr2 == NULL) {fclose(file); return -1;}
+	if (fclose(file) != 0) return -1;
+
+	if ((int)arr2[0] < (int)arr1[0]) return -1;
+
+	qsort(&arr1[1], (int)arr1[0], sizeof(double), ctv);
+	qsort(&arr2[1], (int)arr2[0], sizeof(double), ctv);
+	
+	if (isContains(&arr1[1], (int)arr1[0], &arr2[1], (int)arr2[0]) == 0) {
+		free(arr1);
+		free(arr2);
+		
+		return write("NO");
+	} else {	
+		free(arr1);
+		free(arr2);
+		
+		return write("YES");
+	}
+}