整数 拆分成连续正数序列

 

    x = x*2;
    /* n(1+n)/2 = x*/
    n = (int)sqrt(x); /* a*b =c^2  then a <= c <= b, so n <= sqrt(2x) <= n+1*/
    for(; n >= 2; n--)
    {
        if (x % n)
            continue;
        a1 = x / n - n + 1;
        if (a1 % 2 == 0)
        {
            a1 /= 2;
            show(a1, n);
            showed = 1;
        }
    }

 补充下思想：很简单，先算出可能序列的最大长度n

然后枚举每个n值，每个数位a, a+1, a+2, ..., a+(n-1)，倒过来求a是否存在