C语言训练-求s=a+aa+aaa+aaaa+aa...a的值 （C语言代码）太简洁了 O(n)

解题思路:

                比如给定a=2,n=3:就是2+22+222 = 246；

                                  246 = 6 + 40 + 200 = 3*2 + 2*20 + 1*200 

                那么有：给定a、n： 结果sum = n*a + (n-1)*a*10 + (n-2)*a*10*10 + ……+ 1 *a*10^(n-1)

                我们可以用循环搞定它：sum = 0;

                                                    while(n)

                                                    {

                                                            sum = sum + n*a;

                                                            a = a*10;

                                                            n--;

                                                    }

注意事项:

参考代码:

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <functional> 
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
	int a, n, sum = 0;
	scanf("%d%d", &a, &n);
	while(n)
	{
		sum += n*a;
		a *= 10;
		n--;
	}
	printf("%d\n", sum);
	return 0;
}