Nyoj 1066 co-prime (number theory)

Co-Prime Time limit: +Ms | Memory Limit:65535KB Difficulty:3

Descriptive narrative

This problem are so easy! Can You solve it?

Given a sequence which contains n integers a1,a2......an, your task is to find how many pair (AI, AJ) (I < j) that AI and AJ is co-prime.

Input

There is multiple test cases.
Each test case conatains the Line,the first line contains a single integer n,the second line contains n integers.
All the integers are not greater than 10^5.

Output

for each test case, you should output one of the contains the answer.

Example input

31 2 3

Example output

3

Test instructions: give n positive integers to find out how many of these n numbers are in the number of interconnects.

Analysis:

Fr=aladdin "> The inverse of the mo.

f (d) " Span style= "Font-family:tahoma" >n number of GCD d f (d) n GCD happens to be d

the F (d) =∑f (n) (n d = = 0)

F (d) =∑mu[n/d] * f (n) (n%d = = 0)

The above two formulas are the formulas in Möbius inversion.

So there's a number of pairs that require a mutual-vegetarian. is to ask f (1).

And according to the above formula can be obtained f (1) =∑mu[n] * f (n).

So get mu[] and enumerate n . Mu[i] is the Möbius function of I.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int maxn = 1e5 + 10; typedef long Long Ll;int CNT[MAXN], PRI[MAXN], NUM[MAXN], Pri_num, MU[MAXN], VIS[MAXN], a[maxn];void Mobius (int n)//sieve method to find Mo    Biousse function {pri_num = 0;    memset (Vis, 0, sizeof (VIS));    VIS[1] = mu[1] = 1;            for (int i = 2; I <= n; i++) {if (!vis[i]) {pri[pri_num++] = i;        Mu[i] =-1;            } for (int j = 0; J < Pri_num; J + +) {if (i * pri[j] > N) break;            VIS[I*PRI[J]] = 1;                if (i% pri[j] = = 0) {Mu[i*pri[j]] = 0;            Break        } Mu[i*pri[j]] =-mu[i]; }}}ll get (int x) {return (LL) x * (x-1)/2;}    int main () {Mobius (100005);    int n;        while (~SCANF ("%d", &n)) {int Mmax = 0;            for (int i = 1; I <= n; i++) {scanf ("%d", &a[i]);        Mmax = Max (Mmax, A[i]);        } memset (CNT, 0, sizeof (CNT)); MEmset (num, 0, sizeof (num));        for (int i = 1; I <= n; i++) num[a[i]]++;        for (int i = 1, i <= Mmax; i++) for (int j = i; J <= Mmax; j + = i) cnt[i] + = num[j];        LL ans = 0;        for (int i = 1; I <= Mmax; i++) ans + = Get (Cnt[i]) * Mu[i];    printf ("%lld\n", ans); } return 0;}

Nyoj 1066 co-prime (number theory)