#  因素和求一个自然数 n 的所有因数之和
#  暴力筛复杂度 O (nlogn)
1 2 3 4 5 6 7 8 9 10 11 12 ll f[N],s; int  n;void  init (int  n)     for (int  i=1 ;i<=n;i++)     {         for (int  j=i;j<=n;j+=i)         {             f[j] += i;         }     } } 
+++
#  线性筛参考资料戳此处 p 1 k 1 p 2 k 2 … p n k n p_1^{k_1}p_2^{k_2}…p_n^{k_n} p 1 k 1   p 2 k 2   … p n k n   p 1 p_1 p 1  p 1 2 p_1^2 p 1 2  p 1 k 1 p_1^{k_1} p 1 k 1   p 2 p_2 p 2  p 2 2 p_2^2 p 2 2  p 2 k 2 p_2^{k_2} p 2 k 2   p n p_n p n  p n 2 p_n^2 p n 2  p n k n p_n^{k_n} p n k n   ∑ i = 0 n ∑ j = 0 k i p i j \sum_{i=0}^n\sum_{j=0}^{k_i}p_i^j ∑ i = 0 n  ∑ j = 0 k i   p i j  
#  求三角形面积#  海伦公式S = p ( p − a ) ( p − b ) ( p − c ) S = \sqrt{p(p-a)(p-b)(p-c)}
 S = p ( p − a ) ( p − b ) ( p − c )  
其中
p = a + b + c 2 p = \frac{a+b+c}{2}
 p = 2 a + b + c  
#  秦九韶公式S = 1 4 [ a 2 b 2 − ( a 2 + b 2 − c 2 2 ) 2 ] S = \sqrt{\frac{1}{4}[a^2b^2-(\frac{a^2+b^2-c^2}{2})^2]}
 S = 4 1  [ a 2 b 2 − ( 2 a 2 + b 2 − c 2  ) 2 ]