HDU 5321 Beautiful Set

题目链接
我们能够枚举子集的大小k。求出全部大小为k的子集对答案的贡献。问题就攻克了。
注意到欧拉函数的性质：n=∑φ(d),d|n
莫比乌斯函数性质：∑d|nμ(d)=0n>1
感谢http://blog.csdn.net/u013368721/article/details/47125353

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
typedef long long ll;
typedef long long LL;
const ll inf = 0x3f3f3f3f;
const int mod = 258280327;
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0) {
        x = 1; y = 0; return a;
    } else {
        int d = exgcd(b, a%b, y, x);
        y -= a/b * x;
        return d;
    }
}
int inv(int a)
{
    int x, y, b = mod;
    exgcd(a, b, x, y);
    if (x < 0) x += mod;
    return x;
}
const int N = 102333;
int phi[N];
int fac[N], vf[N];
void init()
{
    fac[0] = 1;
    vf[0] = 1;
    for (int i=1;i<N;i++) {
        phi[i] = i;
        fac[i] = (LL) fac[i - 1] * i % mod;
        vf[i] = inv(fac[i]);
    }
    for (int i=2;i<N;i++) if (phi[i]==i) {
        for (int j=i;j < N; j+=i)
            phi[j] = phi[j] / i * (i - 1);
    }
}
int C(int n, int m)
{
    return (LL) fac[n] * vf[m] % mod * vf[n - m] % mod;
}
int cnt[N];
int n;
int id[N];
bool cmp(int a, int b)
{
    return cnt[a] > cnt[b];
}
// cnt[i] : i 的倍数有多少个
void solve()
{
    int ans1 = 0, ans2 = 0;
    for (int i=1;i<N;i++) {
        int tmp = 0;
        for (int j=1;j<N && cnt[id[j]]>=i;j++) {
            int idx = id[j];
            int t = (LL) C(cnt[idx],i) * phi[idx] % mod;
            tmp = (tmp + t) % mod;
        }
        ans2 = (ans2 + (LL) i * tmp % mod) % mod;
        tmp = (LL) tmp * fac[i] % mod * fac[n-i+1] % mod;
        ans1 = (ans1 + tmp) % mod;
    }
    if (ans1 > ans2) printf("Mr. Zstu %d\n", ans1);
    else {if (ans1 < ans2)
        printf("Mr. Hdu %d\n", ans2);
        else
            printf("Equal %d\n", ans1);
    }
}
int main()
{
    init();
    while (scanf("%d", &n)==1) {
        memset(cnt, 0 ,sizeof cnt);
        for (int i=0;i<n;i++) {
            int x; scanf("%d", &x);
            cnt[x] ++;
        }
        for (int i=1;i<N;i++) {
            id[i] = i;
            for (int j=i+i;j<N;j+=i) {
                cnt[i] += cnt[j];
            }
        }
        sort(id+1, id+N, cmp);
        solve();
    }
    return 0;
}