c -我的for循环添加了+1多余，我不知道为什么

Question

基本上我想做一个程序，通过给定的数组循环，并检查右元素是否比左元素大2倍，如果真插入这两个元素的平均值在中间。之后，它打印出包含插入元素的数组，然后再次遍历数组，计算某个数字出现的次数。我用纸和笔成功地写了所有的代码，把问题写成更小的块，然后用C编码，但问题是，当我输入100个零(100个零)程序打印出数字0重复了200次而不是199次。我不知道为什么。很抱歉代码不好，我现在的任务是善于用笔和纸来解决问题，在我变得体面并发展我的逻辑之后，我会尝试使代码更简单。

Input sample: 
Enter the number of elements: 100
Enter the array: 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
After adding middle element: 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002.33412e-310 
The number is repeated 200 time/s

我的代码

#include <math.h>
#include <stdio.h>
#define EPSILON 0.0001
int main() {
int n, i, j, k, digit, length = 0, digit_array[10] = {0};
double array[200], temp;
do {
printf("Enter number of elements: ");
scanf("%d", &n);
} while (n <= 0 || n >= 101);
printf("Enter elements: ");
length = n;
for (i = 0; i < length; i++)
scanf("%lf", &array[i]);
for (i = 0; i < length; i++) {
temp = array[i] + array[i];
if (fabs(temp - array[i + 1]) < EPSILON) {
for (j = length; j > i + 1; j--)
array[j] = array[j - 1];
array[i + 1] = (array[i] + array[i + 1]) / 2.;
i++;
length++;
}
}
printf("After adding middle element: n");
for (i = 0; i < length; i++)
printf("%g ", array[i]);
for (i = 0; i < length; i++) {
temp = array[i];
digit = ((int)(temp * 10)) % 10;
digit_array[digit]++;
}
printf("n");
for (i = 0; i < 10; i++) {
if (digit_array[i] != 0)
printf("Number %d is repeated %d time/s.n", i, digit_array[i]);
}
return 0;
}

Answer 1

使用两个数组要比不断地移动数组容易得多，也快得多。你只需要:

// Inputs:
//   n: The number of inputs.
//   a: An array of at least n doubles containing the inputs.
//   b: An array of at least n*2-1 doubles that will containing the outputs.
// Outputs:
//   m: The number of outputs.
//   b: An array of at least m doubles containing the outputs.
size_t i = 0;
size_t j = 0; 
double prev = b[j++] = a[i++];
while (i < n) {
double next = a[i];
if (fabs(prev*2 - next) < EPSILON) {   // If a[i-1] exactly equal a[i]*2.
b[j++] = next / 2.0 + prev / 2.0;   // Or: b[j++] = prev * 1.5;
}
prev = b[j++] = a[i++];
}
size_t m = j;

关于prev * 1.5:

average(next, prev)
= ( next + prev ) / 2
= ( prev * 2 + prev ) / 2
= ( prev * 3 ) / 2
= prev * 1.5

包含在适当的函数中:

int f(double *a, size_t n, double **b_ptr, size_t *m_ptr) {
double b = malloc( (n*2-1) * sizeof(double) );  // We need up to this much.
if (b == NULL) {
*b_ptr = NULL;
return 0;
}
size_t i = 0;
size_t j = 0; 
double prev = b[j++] = a[i++];
while (i < n) {
double next = a[i];
if (fabs(prev*2 - next) < EPSILON) {   // If a[i-1] exactly equal a[i]*2.
b[j++] = next / 2.0 + prev / 2.0;   // Or: b[j++] = prev * 1.5;
}
prev = b[j++] = a[i++];
}
b = realloc(b, j * sizeof(double));  // Free the excess. (Optional)
size_t m = j;
*b_ptr = b;
*m_ptr = m;
return 1;
}