提取一个integer的前/后几位

#include <iostream>
using namespace std;

int length(int n){
	int times = 0;
	do {
		n /= 10;
		++times;
	}while(n);
	return times;
}

int main(){
	//get the lastdigits
	int number = 231213;
	int count = 3;                  //extract the last count digits
	int base = 1, result = 0;
	for (int i = 0; i < count; ++i){

		cout << result << " " << number << " " << base << endl;
		result += base * (number % 10);   //原result + 新来的数位 * 10，以保证顺序
		number /= 10;
		base *= 10;
	}
	cout << result << endl;
	
	//get the first digit
	number = 987654321;
	count = 5;	            //extract the first count digits
	int k = length(number) - count;
	for (int i = 0; i < k; ++i){
		number /= 10;
	}
	cout << number << endl;

}

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Summary:对于整数的操作，其实就是 *10 和 /10的合理运用。

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//display the reverse of integers

int reverse(int n){
	int tmp = 1;             //区别
	int sum = 0;
	for (;n!=0;){
		tmp = n % 10;
		n = n /10;
		sum = sum*10 + tmp;  //原sum * 10 + 新来的数位，则逆序
 	}
	return sum;
}