1. Problem description

problem:

Give a positive integer with no more than 5 digits, and require:

① Find out how many digits it is;

② Output each digit separately;

③Output each digit in reverse order

Examples:

123 is a 3-digit number, the hundreds, tens, and ones digits are 1, 2 and 3 respectively; the output in reverse order is 321;

enter:

123

Output:

1 2 3

321

2. Problem solving

problem analysis:

(1) Find the number of digits of a positive integer

Method 1: Divide a positive integer by 10, each time the integer can be divided by one, and the count is increased by one, until the integer is 0;

For example: 123/10=12 count count=1; 12/10=1 count count=2; 1/10=0 count count=3, the quotient is 0, the end, the integer 123 is a 3-digit number.

programming:


//求正整数的位数
int figure(int num) {
	int count = 0;
	if (num == 0) {
		count = 1;
	}
	else {
		while (num) {
			num /= 10;
			count++;
		}
	}
	return count;
}

Method 2: Use the function log10(x)+1

log() function introduction:

Header file : <math.h>

Function prototype : double log(double x);

Function : Find the logarithm based on natural numbers

Parameters : double x is a true number and must be greater than 0

Return value : Returns the logarithm based on the natural number

Formula : loge x = b

Note : The natural number e is a constant 2.71828

When the base is changed to 10, the number of digits can be solved. We know that log10(10)=1, the closed interval [1-9] is a 1-digit integer; log10(100)=2, the closed interval [10-99] is a 2-digit integer, so the number of integers to be solved is log10( x)+1 ;

int figure2(int num) {
	if (num == 0) {
		return 1;
	}
	else {
		return ((int)log10(num)) + 1;//由于log函数返回类型是double，所以使用（int)进行强制转换
	}
}

(2) Find the number on each digit of an integer

The remainder of an integer is 10 (%10) to get the number on the last digit of the integer, for example: 123%10=3 ;

Dividing an integer by 10 (/10) can eliminate the last digit of the integer, for example: 123/10=12 (because of an integer/integer, the result is still an integer, which is equivalent to rounding the result)

//求整数各位的数字
/*
返回类型为：整型数组指针（返回的是存放各位数字的数组起始地址）
num为待处理数字，数组a为存储数组
*/
int* figure_num(int num,int a[5]) {
	int i = 0;
	while(num){
		a[i] = num % 10;
		num /= 10;
		i++;
	}
	return a;
}

int main() {
	int num = 123;
	int len = figure2(num);//计算整数位数
	int arr[5];
	int* p = arr;
	p = figure_num(num,arr);//返回存储地址
	for (int i = len-1; i >=0; i--) {
		printf("%d ", arr[i]);//由高位向低位依次输出
		
	}
}

(3) Reverse order output

It's relatively simple. When directly looping output, press 0~len-1 to output in turn.

Infer other things: What if you want to output the reversed integer of an integer?

//逆序数
int reverse_num(int num) {
	int res = 0;
	while (num) {
		res = res * 10 + num % 10;
		num /= 10;
	}
	return res;
}

[C Programming] Simple programming exercises-(3) Find the number of positive integers, each digit and the reverse integer

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