“一步千里”之数组找数

这里我给出另一种代码：

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

/*
 * to find a target number in a special array 
 * */
int find(int arr[], int n, int start,  int target) {
    int distance = fabs(target - arr[start]);    // to accelerate the search
    if(start < n) {
    if(arr[start + distance] == target) {
        printf("%d has been found in the position of %d\n",
            target, start + distance);
        return 1;
    }
    else return find(arr, n, start + distance, target);
    }
    else return 0;
}
int main() {
    const int n = 10;
    int arr[n];
    int start = 0;
    int target;

    int i;
    for(i = 0;i < n;i++) {
    printf("the %d number is  = ", i);
    scanf("%d", &arr[i]);
    }

    printf("the number should be search is:");
    if(scanf("%d", &target) != 1) {
    perror("Input Error!\n");
    exit(1);
    }
   
    if(find(arr, n, start, target) == 0) {
    printf("can not find %d\n", target);
    }
    return 0;
}

测试：