谢谢了!


> sum = a1+a2+...+an
> sum = an+...+a2+a1
> 
> 2sum = n(a1+an)
> 又an = a1+ (n-1)
> 
> 2sum = n(2a1+n-1)
> 
> eg1: 
> sum=15;
> 2sum = 30 = n(2a1+n-1)
> 
> 把30 分解成 a*b的形式:
> 
>  2sum = n * (2*a1+n-1)
>    30 = 1 * 30     ==> 30 = 2*a1+1-1 ==> a1=15 (n=1)
>       = 2 * 15    ==> 15 = 2*a1+2-1 ==> a1=7  (n=2)
>       = 3 * 10    ==> 10 = 2*a1+3-1  ==> a1=4  (n=3)
>       = 5 * 6     ==>  6 = 2*a1+5-1  ==>  a1=1  (n=5)
>   
> eg2：
> sum = 10,
>  2sum = 20 = n(2a1+n-1)
> 
> 把20 分解成 a*b的形式:
> 
>   2sum = n * (2*a1+n-1)
>    20 = 1 * 20     ==> 20 = 2*a1+1-1 ==> a1=10 (n=1)
>       = 2 * 10    ==> 10 = 2*a1+2-1 ==> 2*a1 = 9, 不存在
>       = 4 * 5    ==> 5 = 2*a1+4-1  ==> a1=1  (n=4)
>   
> 
>