Noj 2079 Prime (Möbius inversion)

Prime

Time limit (normal/java): 1000ms/3000ms Run memory limit: 65536KByte
Total Submissions: 267 tested by: 11

Contest description
For a given number of N, 22 mutually exclusive logarithm is obtained. Mutual exclusion refers to the two number of greatest common divisor is 1

Input
First behavior Example number T (t<=5)
For each example, the first behavior is an integer n (2=<n<=10^5), which represents the number of numbers.
The next line contains the number of N, A1,a2,..., an (1<=ai<=10^5)

Output
For each sample, output the answer on one line.

Sample input
1
2
2 3

Sample output
1

Title Link: http://acm.njupt.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2079

Title analysis: First preprocessing at the Möbius function of 1e5, because mob[i]=1 indicates that I is the product of an even number of factors, mob[i]=-1 that I is the product of the odd number of factors, mob[i]=0 represents other
Back to this problem, first for the total collection I arbitrarily take out two number of cases of C (n,2) that is n * (n-1)/2, and then subtract the case, that is, GCD 2 gcd 3, here is a problem, because if GCD 6, then it was reduced two times, so to let out, That is to take the total situation-gcd only by a factor of one element of the situation +GCD only two factors constitute the case ... For each set of GCD values we can record with a num array. Then we can see that the Möbius function is powerful, the corresponding positive and negative number is actually a mob array, such as the calculation of GCD to 2, mob[2]==-1, so minus num[2] * (num[2]-1)/2,3 time is also reduced, 6 is added, Coincides with the definition of the mob function.

Note: Because the Möbius function is an integrable function, it is possible to use a linear sieve to derive

#include <cstdio> #include <cstring> #include <algorithm> #define LL long longusing namespace Std;int    Const MAX = 1e5 + 5;int P[max], Cnt[max], Num[max], Mob[max];bool prime[max];int pnum, MA, n;void Mobius ()//solution Möbius function {    Pnum = 0;    MOB[1] = 1;    Memset (Prime, true, sizeof (prime));            for (int i = 2; i < MAX; i++) {if (Prime[i]) {p[pnum + +] = i;        Mob[i] =-1;            } for (int j = 0; J < pnum && I * p[j] < MAX; J + +) {Prime[i * P[j]] = false;                if (i% p[j] = = 0) {Mob[i * p[j]] = 0;            Break        } Mob[i * P[j]] =-mob[i];    }}}ll cal () {ll ans = (LL) n * (n-1)/2;        for (int i = 2; I <= ma; i++) {num[i] = 0;   for (int j = i; J <= Ma; j + = i) num[i] + = cnt[j]; Gets the number of elements of the set GCD to i} for (int i = 2; I <= ma; i++) ans + = (ll) mob[i] * num[i] * (Num[i]-1)/2; return ans;}    int main () {Mobius ();    int T;    scanf ("%d", &t);        while (t--) {memset (CNT, 0, sizeof (CNT));        Ma = 0;        scanf ("%d", &n);            for (int i = 0; i < n; i + +) {int tmp;            scanf ("%d", &tmp);            CNT[TMP] + +;        MA = max (MA, TMP);    } printf ("%i64d\n", Cal ()); }}

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Noj 2079 Prime (Möbius inversion)