Hdu 5072 Coprime (number theory), hducoprime

Hdu 5072 Coprime (number theory), hducoprime

Link: hdu 5072 Coprime

Given the number of N, ask how many 3 tuples can be selected, or [(a, B) = (B, c) = (a, c) = 1] or [(a, B) =1 and (a, c) =1 and (B, c) =

1].

Solution: You can change the angle of this question to think of three numbers as three sides of a triangle. The mutual quality means that the color of the side is 1. Otherwise, the value is 0.

Number of triangles with the same color as the three sides.

The total number of triangles can be obtained. If the number of triangles with different three edges is obtained, subtract them.

Enumerate the vertex, and then determine the number of non-satisfied triangles formed by the triangle formed by the vertex, that is, select one in 1 and one in 0. This is not enough

The triangle is calculated twice, ans/2.

Then the problem is converted to finding the number of each number of mutual or non-mutual quality, dividing each number difference by quality factor, and counting the number of each number containing quality factor x,

When calculating each number, you can use the exclusion principle.

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 1e5;int np, pri[maxn + 5], vis[maxn + 5];int k, fac[maxn + 5];int v[maxn + 5];vector<int> g[maxn + 5];void prime_table(int n) {    np = 0;    for (int i = 2; i <= n; i++) {        if (vis[i])            continue;        pri[np++] = i;        for (int j = i * 2; j <= n; j += i)            vis[j] = 1;    }}void div_fac(int n, int& cnt) {    cnt = 0;    for (int i = 0; i < np && pri[i] * pri[i] <= n; i++) {        if (n % pri[i] == 0) {            fac[cnt++] = pri[i];            while (n % pri[i] == 0)                n /= pri[i];        }    }    if (n != 1)        fac[cnt++] = n;}void dfs (int u, int d, int idx) {    if (d >= v[idx]) {        if (u != 1)            g[idx].push_back(u);        return;    }    dfs(u, d + 1, idx);    dfs(u * fac[d], d + 1, idx);}void init () {    np = 0;    memset(vis, 0, sizeof(vis));    prime_table(maxn);    for (int i = 2; i <= maxn; i++) {        div_fac(i, v[i]);        dfs(1, 0, i);    }}int N, f[maxn + 5], c[maxn + 5];void input () {    int x;    scanf("%d", &N);    memset(c, 0, sizeof(c));    memset(f, 0, sizeof(f));    for (int i = 0; i < N; i++) {        scanf("%d", &x);        c[x]++;    }    for (int i = 2; i <= maxn; i++) {        if (c[i] == 0)            continue;        for (int j = 0; j < g[i].size(); j++)            f[g[i][j]] += c[i];    }}typedef long long ll;int count (int n) {    int ret = 0;    for (int i = 0; i < g[n].size(); i++) {        int k = g[n][i];        if (v[k]&1)            ret += (f[k] - 1);        else            ret -= (f[k] - 1);    }    return ret;}ll solve () {    ll ret = 0;    for (int i = 2; i <= maxn; i++) {        if (c[i] == 0)            continue;        ll k = count(i);        ret += k * (N - 1 - k);    }    ll a = N;    ll sum = 1LL * a * (a - 1) * (a - 2) / 6;    return sum - ret / 2;}int main () {    init();    int cas;    scanf("%d", &cas);    while (cas--) {        input();        printf("%I64d\n", solve());    }    return 0;}