Hdu5514frogs (2015 Shenyang Division onsite)

This problem can easily be converted to a tolerance counting issue. The enumeration counting method with exponential complexity obviously hangs and can only be considered in other ways.

First the a[i] is replaced with GCD (A[i], M), and the sort is weighed to get an approximate set of M, and M is not more than 1e9, so all the approximate number of M is around 1000.

Assuming that the array a contains only 2, 32 elements, then it is clear that the answer should be f (2) + F (3)-F (6) and F (i) for the case of only I.

The coefficients of 2 and 3 are the coefficients of 1,6 are-1, the remaining coefficients are 0.

According to the principle of repulsion: for a particular approximate number, if there is a subset of array a whose LCM is the approximate, then if the set of elements is even, contribute 1, if it is odd, contribute 1.

The final answer must be synthesized by all approximations with a value of { -1,0,1}.

Therefore, we consider the number of the first m of array A, and its corresponding coefficient distribution can be determined.

When adding the number of m+1, it is obvious that the new set must be obtained by a subset of the original set and a new element, considering all the approximate k of M, with the corresponding coefficients op (k)

If OP (k) is-1, it means that the number of even subsets is 1 more than the number of odd subsets, and if we add a new element x, it just makes the odd subset number 1 more than an even number, but at this point the contribution

Not to K, but to the LCM (k, X), which is represented by a formula:

OP (LCM (k, x)) + = OP (k) * (-1).

The rest of the situation is similar.

The final answer is ans = sigma (OP (i) * f (i)), I | M.

This allows us to perform a maximum of 1000 traversal so that the number of elements is added one and the total complexity is approximately o (1e6).

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <map>5#include <string>6#include <vector>7#include <Set>8#include <cmath>9#include <ctime>Ten#include <cassert> One #defineLson (U << 1) A #defineRson (U << 1 | 1) - #defineCLS (i, J) memset (i, J, sizeof i) - #pragmaComment (linker, "/stack:102400000,102400000") the using namespacestd; - typedef __int64 LL; - Const DoubleEPS = 1e-6; - Const DoublePI = ACOs (-1.0); + Const intMAXN = 1e4 +Ten; - Const intMAXM = 4e4 +Ten; + Const intINF =0x3f3f3f3f; A Constll linf =0x3fffffffffffffff; at Constll mod = 1e9 +7; -  -Map<ll,int>MAPI; - intN; - ll A[maxn], m; - intPRIME[MAXM], K; in BOOLVIS[MAXM]; - ll FACT[MAXN], NF; to intCNT[MAXN]; + ll TABLE[MAXN], NT; - intOP[MAXN]; the intOP1[MAXN]; * ll ans; $ Panax Notoginsengll GCD (ll A, ll b) {return!b? A:GCD (b, a%b); } -  the voidinit () { +CLS (Vis,0); AK =0; the      for(inti =2; i < MAXM; i++){ +         if(Vis[i])Continue; -prime[k++] =i; $          for(intj = i *2; J < Maxm; J + = i) vis[j] =1; $     } - } -  the voidDfsintNext, ll num) { -     if(Next >=NF) {Wuyitable[nt++] =num; theMapi[num] = NT-1; -         return; Wu     } -ll tem =num; About      for(inti =0; I <= Cnt[next]; i++){ $DFS (next +1, TEM); -TEM *=Fact[next]; -     } - } A  + ll F (ll num) { thell tem = (M-1) /num; -     returnTEM * (TEM +1) /2*num; $ } the  the voidfactorize () { theNF =0; theLL M1 =m; -     intMID = (int) sqrt (m); in      for(inti =0; Prime[i] <= mid; i++){ the         if(M% prime[i])Continue; thefact[nf++] =Prime[i]; AboutCNT[NF-1] =0; the          while(m% prime[i] = =0) ++CNT[NF-1], M/=Prime[i]; theMID = (int) sqrt (m); the     } +     if(M! =1) fact[nf++] = m, cnt[nf-1] =1; -m =M1; the }Bayi  the ll Solve () { the mapi.clear (); -     intN1 =0; -      for(inti =0; I < n; i++){ thell tem = A[i]%m; the         if(!tem)Continue; thea[n1++] =gcd (M, TEM); the     } -n =N1; theSort (A, A +n); then = Unique (A, a + N)-A; theCLS (Vis,0);94      for(inti =0; I < n; i++){ the          for(intj = i +1; J < N; J + +){ the             if(A[i]% a[j] = =0) Vis[i] =1; the             Else if(A[j]% a[i] = =0) Vis[j] =1;98         } About     } -N1 =0;101      for(inti =0; I < n; i++)if(!vis[i]) a[n1++] =A[i];102n =N1;103Sort (A, A +n);104 factorize (); theNT =0;106Dfs0,1);107CLS (OP,0);108Ans =0;109      for(inti =0; I < n; i++){ theCLS (OP1,0);111          for(intj =0; J < NT; J + +){ thell tem = A[i]/gcd (A[i], table[j]) *Table[j];113             if(TEM >= m)Continue; theOp1[mapi[tem]] + =-Op[j]; the         } the++Op1[mapi[a[i]];117          for(intj =0; J < NT; J + +) Op[j] + =Op1[j];118     }119    //for (int i = 0; i < NT; i++) printf ("%d", Op[i]); -    //puts ("");121      for(inti =0; I < NT; i++) ans + = op[i] *f (table[i]);122     returnans;123 }124  the intMain () {126    //freopen ("In.txt", "R", stdin);127     intT, Kase =0; - init ();129scanf"%d", &T); the      while(t--){131scanf"%d%i64d", &n, &m); the          for(inti =0; I < n; i++) scanf ("%i64d", &a[i]);133ll ans =solve ();134printf"Case #%d:%i64d\n", ++Kase, ans);135     }136     return 0;137}

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Hdu5514frogs (2015 Shenyang Division onsite)