HDU 4675 GCD of Sequence（莫比乌斯反演 + 打表注意事项）题解

题意：

给出\(M\)和\(a数组\)，询问每一个\(d\in[1,M]\)，有多少组数组满足：正好修改\(k\)个\(a\)数组里的数使得和原来不同，并且要\(\leq M\)，并且\(gcd(a_1,a_2,\dots,a_n)=d\)。

思路：

对于每一个\(d\)，即求\(f(d)\)：修改\(k\)个后\(gcd(a_1,a_2,\dots,a_n)=d\)的对数。

那么假设\(F(d)\)：修改\(k\)个后\(gcd(a_1,a_2,\dots,a_n)\)是\(d\)倍数的对数。故：

\[f(k) = \sum_{k|d}\mu(\frac{d}{k})F(d)
\]

打表求\(F(d)\)即可。假设\(num[d]\)为\(a\)中是\(d\)倍数的数量，则

\[F(d)=(\frac{M}{d})^{n-num[d]}*C_{num[d]}^{k-(n-num[d])}*(\frac{M}{d}-1)^{k-(n-num[d])}
\]

然后\(nlogn\)打出\(num\)数组即可。

思考：

这样的打表法是\(nlogn\)的：

证明 O(n/1+n/2+…+n/n)=O(nlogn)

for(int i = 1; i <= n; i++){
     scanf("%d", &a[i]);
     num[a[i]]++;
 }
 for(int i = 1; i <= m; i++){
     for(int j = i + i; j <= m; j += i){
         num[i] += num[j];
     }
 }

这样是\(n\sqrt n\)的

for(int i = 1; i <= n; i++){
     scanf("%d", &a[i]);
     for(int j = 1; j <= sqrt(a[i]); j++){
	     if(a[i] % j == 0){
		     num[j]++;
		     if(j * j != a[i]) num[a[i] / j]++;
	     }
     }
 }

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1000000007;
using namespace std;

int mu[maxn], vis[maxn];
int prime[maxn], cnt;
ll fac[maxn], inv[maxn];
ll ppow(ll a, ll b){
    ll ret = 1;
    while(b){
        if(b & 1) ret = ret * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ret;
}
void init(int n){
    memset(vis, 0, sizeof(vis));
    memset(mu, 0, sizeof(mu));
    cnt = 0;
    mu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && prime[j] * i <= n; j++){
            vis[prime[j] * i] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }

    fac[0] = inv[0] = 1;
    for(int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
    inv[n] = ppow(fac[n], MOD - 2);
    for(int i = n - 1; i >= 1; i--) inv[i] = (i + 1LL) * inv[i + 1] % MOD;
}
ll C(int n, int m){
    return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
int num[maxn], a[maxn];
//num[i]:是i的倍数的个数
ll F[maxn], f[maxn];
int main(){
    init(3e5);
    int n, m, k;
    while(~scanf("%d%d%d", &n, &m, &k)){
        memset(num, 0, sizeof(num));
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            num[a[i]]++;
        }
        for(int i = 1; i <= m; i++){
            for(int j = i + i; j <= m; j += i){
                num[i] += num[j];
            }
        }
        for(int i = 1; i <= m; i++){
            int no = n - num[i];
            if(no > k) F[i] = 0;
            else{
                F[i] = ppow(m / i, no) * C(num[i], k - no) % MOD * ppow(m / i - 1, k - no) % MOD;
            }
        }

        for(int i = 1; i <= m; i++){
            f[i] = 0;
            for(int j = i; j <= m; j += i){
                f[i] += mu[j / i] * F[j];
                f[i] %= MOD;
            }
            printf("%lld%c", (f[i] % MOD + MOD) % MOD, i == m? '\n' : ' ');
        }

    }
    return 0;
}