recursive inverse

three lines of code

inv[1] = 1;
for (int i = 2; i<MAXN; i++)

inv [i] = inv [MOD% i] * (MOD-MOD / i)% MOD;

Just to demonstrate the feasibility:

• Just prove: inv[i]*i=1mod MOD
• 可 证: inv [MOD% i] * (MOD-MOD / i)% MOD * i = 1 mod MOD
• Let MOD = i*k+b;
• 左边 = inv [MOD% i] * (MOD% i) (MOD * i-MOD)% MOD / (MOD% i) mod MOD
•        = (MOD-MOD / i) * i% MOD / (MOD% i) mod MOD
  
•        = (MOD-k) * in% MOD / b against MOD
  
•        = (MOD * i + (b-MOD))% MOD / b mod MOD
  
•        = b / b% MOD mod MOD 
  
•        = 1 mod MOD
  
• certified