Question

出于某种原因，何时

SIZEOF_A = 6
SIZEOF_B = 4
set A = {0, 1, 2, 3, 4, 5, 6}
set B = {1, 10, 11, 12}

交叉点都很棘手（1048832）。这是为什么？

#include "usefunc.h"

/* #define SIZEOF_A 2
#define SIZEOF_B 4 */

typedef enum {
        a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
} set;

void dispSet(set numbers[], int size_numbers) {
  int i;
  printf("[ ");
  for (i = 0; i < size_numbers-1; i++) {
    printf("%d, ", numbers[i]);
  }
  printf("%d ]", numbers[size_numbers-1]);
  printf("\n");
}

int in_arr(int A, set B[], int B_size) {
    int res = 0;
    int counter;
    for (counter = 0; counter < B_size; counter++) {
        if (A == B[counter]) {
            res = 1;
            break;
        }
    }
    return res;
}

int arr_in_arr(set smaller[], int smaller_size, set bigger[], int bigger_size) {
    int res = 1;
    int counter;
    for (counter = 0; counter < smaller_size; counter++) {
        if (in_arr(smaller[counter], bigger, bigger_size)) {
            continue;
        }
        else {
            res = 0;
            break;
        }
    }
    return res;
}

int size_c(set arr1[], int arr1_size, set arr2[], int arr2_size) {
    int i;
    int newsize = 0;
    for (i = 0; i < arr1_size; i++) {
        if (!in_arr(arr1[i], arr2, arr2_size)) {
            newsize++;
        }
    }
    for (i = 0; i < arr2_size; i++) newsize++;
    return newsize;
}

int size_d(set arr1[], int arr1_size, set arr2[], int arr2_size) {
    int i;
    int SIZEOF_D = 0;
    for (i = 0; i < arr1_size; i++) if (in_arr(arr1[i], arr2, arr2_size)) SIZEOF_D++;
    return SIZEOF_D;
}

int Join(set arr1[], int arr1_size, set arr2[], int arr2_size, set arr3[], int arr3_size) {
    int i, j = 0;
    for (i = 0; i < arr1_size; i++) arr3[i] = arr1[i];

    for (i = 0; i < arr2_size; i++) {
        if (!in_arr(arr2[i], arr3, arr3_size)) {
            arr3[j+arr1_size] = arr2[i];
            j++;
        }
    }
}

int Intersection(set arr1[], int arr1_size, set arr2[], int arr2_size, set arr3[], int arr3_size) {
    int i, j = 0;
    for (i = 0; i < arr1_size; i++) {
        if (in_arr(arr1[i], arr2, arr2_size)) arr3[j] = arr1[i];
        j++;
    }
}

int getSet(set arr[], int size) {
    int i;
    printf("Starting with a=0, enter a number that corresponds to a letter.\n");
    for (i = 0; i < size; i++) {
        printf("NUM: ");
        arr[i] = GetInteger();
    }
}

int main () {
    printf("size of a? "); int SIZEOF_A = GetInteger();
    printf("size of b? "); int SIZEOF_B = GetInteger();
    set A[SIZEOF_A]; printf("\nA:\n"); getSet(A, SIZEOF_A);
    set B[SIZEOF_B]; printf("B:\n"); getSet(B, SIZEOF_B);
    int SIZEOF_C = size_c(A, SIZEOF_A, B, SIZEOF_B);
    int SIZEOF_D = size_d(A, SIZEOF_A, B, SIZEOF_B);
    printf("size of D: %d", SIZEOF_D);
    set C[SIZEOF_C]; set D[SIZEOF_D];
    Join(A, SIZEOF_A, B, SIZEOF_B, C, SIZEOF_C);
    Intersection(A, SIZEOF_A, B, SIZEOF_B, D, SIZEOF_D);
    printf("\nFor the sets\n");
    printf("A: "); dispSet(A, SIZEOF_A); 
    printf("B: "); dispSet(B, SIZEOF_B);
    printf("\n"); printf("union A,B: "); dispSet(C, SIZEOF_C);
    printf("intersection A,B: "); dispSet(D, SIZEOF_D);
    printf("%s\n\n", (arr_in_arr(A, SIZEOF_A, B, SIZEOF_B) == 1)?"B contains A":"B does not contain A");
}

Answer 1

在交点中，您的if语句应为：

if (in_arr(arr1[i], arr2, arr2_size)) {
    arr3[j] = arr1[i];
    j++;
}

除非找到匹配项，否则不要增加j。

两个数组的交集 - C.

1 个答案: