Hdu 4983 Goffi and GCD (number theory), hdugoffi

Hdu 4983 Goffi and GCD (number theory), hdugoffi

Link: hdu 4983 Goffi and GCD

Question: How many pairs of tuples meet the formula in the question.

Solution:

• When n = 1 or k = 2: the answer is 1.
• K> 2: the answer is 0 (n = 1)
• K = 1: Calculate the factor of n, so that the factor k = gcd (n −a, n, then the gcd of the other side (n −b, n) = nk can satisfy the multiplication of n, and the number of a that satisfies k = gcd (n−a, n) is round (n/s ), euler's algorithm with o (n random √)

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1e5;const int MOD = 1e9+7;typedef long long ll;ll ans;int N, K, M;int np, pri[maxn+5], vis[maxn+5];int nf, fact[maxn+5], coun[maxn+5];void prime_table (int n) {    np = 0;    memset(vis, 0, sizeof(vis));    for (int i = 2; i <= n; i++) {        if (vis[i])            continue;        pri[np++] = i;        for (int j = 2 * i; j <= n; j += i)            vis[j] = 1;    }}void div_factor(int n) {    nf = 0;    for (int i = 0; i < np; i++) {        if (n % pri[i] == 0) {            coun[nf] = 0;            while (n % pri[i] == 0) {                coun[nf]++;                n /= pri[i];            }            fact[nf++] = pri[i];        }    }    if (n != 1) {        coun[nf] = 1;        fact[nf++] = n;    }}int euler_phi(int n) {    int m = (int)sqrt((double)n+0.5);    int ret = n;    for (int i = 2; i <= m; i++) {        if (n % i == 0) {            ret = ret / i * (i-1);            while (n%i==0)                n /= i;        }    }    if (n > 1)        ret = ret / n * (n - 1);    return ret;}ll add (int s) {    ll a = euler_phi(N/s);    ll b = euler_phi(s);    ll ret = a * b * 2;    if (s == N / s)        ret /= 2;    return ret;}void dfs (int s, int d) {    if (s > M)        return;    if (d == nf) {        ans = (ans + add(s)) % MOD;        return;    }    for (int i = 0; i <= coun[d]; i++) {        dfs(s, d+1);        s *= fact[d];    }}int solve () {    ans = 0;    M = (int)sqrt((double)N);    div_factor(N);    dfs (1, 0);    return ans % MOD;}int main () {    prime_table(maxn);    while (scanf("%d%d", &N, &K) == 2) {        if (N == 1)             printf("1\n");        else if (K > 2)            printf("0\n");        else if (K == 2)            printf("1\n");        else            printf("%d\n", solve());    }    return 0;}