HDU 5321 Beautiful Set 容斥 （看题解）

Posted 2022-07-20 cjlhy

HDU 5321

感觉有点抗拒这种题目， 看到就感觉自己不会写，其实就是个沙雕题， 感觉得找个时间练练这种题。

g[ i ] 表示gcd为 i 的倍数的方案数， f[ i ] 表示gcd为 i 的方案数， 然后先算g[ i ]然后直接容斥。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 258280327;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) a += b; if(a >= mod) a -= mod;
template<class T, class S> inline void sub(T &a, S b) a -= b; if(a < 0) a += mod;
template<class T, class S> inline bool chkmax(T &a, S b) return a < b ? a = b, true : false;
template<class T, class S> inline bool chkmin(T &a, S b) return a > b ? a = b, true : false;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, a[N], cnt[N];

int F[N], Finv[N], inv[N];
int f[N], g[N];


void prepare() 
    F[0] = Finv[0] = inv[1] = 1;
    for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod;
    for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;


inline int A(int n, int m) 
    if(n < 0 || n < m) return 0;
    return 1LL * F[n] * Finv[n - m] % mod;


inline int C(int n, int m) 
    if(n < 0 || n < m) return 0;
    return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;


int solve1() 
    for(int i = 1; i <= 100000; i++) 
        g[i] = 0;
        for(int j = 1; j <= cnt[i]; j++) 
            add(g[i], 1LL * A(cnt[i], j) * F[n - j + 1] % mod);
        
    
    int ans = 0;
    for(int i = 100000; i >= 1; i--) 
        f[i] = g[i];
        for(int j = i + i; j <= 100000; j += i) 
            sub(f[i], f[j]);
        
        add(ans, 1LL * i * f[i] % mod);
    
    return ans;



int solve2() 
    for(int i = 1; i <= 100000; i++) 
        g[i] = 0;
        for(int j = 1; j <= cnt[i]; j++) 
            add(g[i], 1LL * j * C(cnt[i], j) % mod);
        
    
    int ans = 0;
    for(int i = 100000; i >= 1; i--) 
        f[i] = g[i];
        for(int j = i + i; j <= 100000; j += i) 
            sub(f[i], f[j]);
        
        add(ans, 1LL * i * f[i] % mod);
    
    return ans;


void init() 
    for(int i = 1; i < N; i++) 
        cnt[i] = 0;
    


int main() 
    prepare();
    while(scanf("%d", &n) != EOF) 
        init();
        for(int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
            cnt[a[i]]++;
        
        for(int i = 1; i <= 100000; i++) 
            for(int j = i + i; j <= 100000; j += i) 
                cnt[i] += cnt[j];
            
        

        int ans1 = solve1();
        int ans2 = solve2();

        if(ans1 == ans2) 
            printf("Equal %d\n", ans1);
        
        else if(ans1 > ans2) 
            printf("Mr. Zstu %d\n", ans1);
        
        else 
            printf("Mr. Hdu %d\n", ans2);
        
    
    return 0;


/*
*/