Ultra-large number modulus of UVA 10692 Huge Mods

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Test instructions is not difficult to understand, because the index range is too large, so I think it is necessary to use power Dafa: ab% C = ab% phi (c) + phi (c)% c (B > Phi (c)) ? Later found that really need to use, and because it has a lot of heavy index, so need DFS, deep search to the last layer before returning, each up a layer to return with a power formula to deal with the index, and then this layer with the same principle to deal with the current layer of the value of the modulus, and back up a layer. Euler function preprocessing can, the end of the problem is also a bit of people, I am using input hanging to deal with.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 Const intM =10006;6 7 intPhi[m] = {0,1,};8InlinevoidInitintn = M-3) {9      for(inti =2; I <= N; ++i)Ten         if(!Phi[i]) One          for(intj = i; J <= N; J + =i) { A             if(!phi[j]) phi[j] =J; -PHI[J] = phi[j]/I * (i-1); -         } the } -  -#include <cctype> -InlineintReadint&x) { +x =0; -     CharCH =GetChar (); +      while(!isdigit (CH) && ch! ='#') ch =GetChar (); A     if(ch = ='#')return 0; at      while(IsDigit (ch)) { -x = x *Ten+ (CH-'0'); -CH =GetChar (); -     } -     return 1; - } in  - intQuick_mod (intAintBintm) { to     intres =1; +      while(b) { -         if(B &1) Res = res * A%m; theA = a * a%m; *b >>=1; $     }Panax Notoginseng     returnRes; - } the  + intm,n,a[ -]; A  the //The ID is the array subscript for the current layer, and the mod is the modulus for the current layer, + //It is known from the Power formula that the mod is the Euler function of the MoD that passes the current layer, that is, the value of Phi[mod] each time the next level of the argument is passed. - //DFS up one layer returns the exponent that is processed by the exponentiation formula, and more see the code. $ intDfsintIdintMoD) { $     if(id = =N) { -         if(A[id] > MoD)returnA[id]% mod +MoD; -         Else    returnA[id]; the     } -     intPOW = DFS (id +1, Phi[mod]);Wuyi  the     intMul =1, C = a[id], num =POW; -  Wu     //because ignoring the C <= MoD led to an intermediate data overflow that caused me to TLE several times, and T had to know the truth, -     //still thinking is not the complexity of the wrong, so I step by step to the painful to debug t.t About      while(Num && mul <= mod && c <=MoD) { $         if(Num &1) Mul *=C; -C *=C; -Num >>=1; -     } A     if(Num && (mul > MoD | | c > MoD))returnQuick_mod (A[id], pow, MoD) +MoD; +     Else    returnMul; the } -  $ intMain () { the     intCase =0; the init (); the      while(Read (m)) { the read (n); -          for(inti =1; I <= N; ++i) in read (a[i]); theprintf"Case #%d:%d\n", ++case, DFS (1, m)%m); the     } About     return 0; the}

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For a long time did not do number theory, it is really cool feeling! Although it is very difficult, although I have more than n XX theorem will not, but I will not give up this so attractive branch of mathematics, think of the original ACM has a lot of reason also because she ~

Ultra-large number modulus of UVA 10692 Huge Mods