hdu 4675 GCD of Sequence（计数）

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题目链接：hdu 4675 GCD of Sequence

代码

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 300000;const int mod = 1e9 + 7;int fac[maxn + 5], inv[maxn + 5];vector<int> P[maxn + 5];int pow_mod(int x, int n) {    if (n < 0) return 0;    int ret = 1;    while (n) {        if (n&1) ret = 1LL * ret * x % mod;        x = 1LL * x * x % mod;        n >>= 1;    }    return ret;}void presolve (int n) {    for (int i = 0; i <= n; i++) P[i].clear();    for (int i = 1; i <= n; i++) {        for (int j = i; j <= n; j += i)            P[j].push_back(i);    }    fac[0] = inv[0] = 1;    for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i-1] * i % mod;    inv[n] = pow_mod(fac[n], mod-2);    for (int i = n-1; i; i--) inv[i] = 1LL * inv[i+1] * (i+1) % mod;}int C(int n, int k) {    if (n < 0 || k < 0 || k > n) return 0;    return 1LL * fac[n] * inv[k] % mod * inv[n-k] % mod;}int N, M, K, cnt[maxn + 5], ans[maxn + 5];void init () {    int x;    memset(cnt, 0, sizeof(cnt));    for (int i = 1; i <= N; i++) {        scanf("%d", &x);        for (int j = 0; j < P[x].size(); j++)            cnt[P[x][j]]++;    }}int main () {    presolve(maxn);    while (scanf("%d%d%d", &N, &M, &K) == 3) {        init();        for (int i = M; i; i--) {            int k = M / i;            ans[i] = 1LL * C(cnt[i], N-K) * pow_mod(k-1, cnt[i]-N+K) % mod * pow_mod(k, N-cnt[i]) % mod;            for (int j = 2; j * i <= M; j++)                ans[i] = (ans[i] - ans[j*i]) % mod;            ans[i] = (ans[i] + mod) % mod;        }        for (int i = 1; i <= M; i++)            printf("%d%c", ans[i], i == M ? '\n' : ' ');    }    return 0;}

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