数列的 GCD [计数问题]

$$数列的GCD$$

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$\mathcal{D}escription$

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$\mathcal{S}olution$

由于有 $K$ 个位置两数列不相同, 于是 使用 {an}\{a_n\}{an​} 构造 {bn}\{b_n\}{bn​},

枚举 $d$, 即假设 $d$ 是已知量,
设 {an}\{a_n\}{an​} 有 $x$ 个位置不能整除 $d$,
由于$条件2$的限制, 这 $x$ 个位置需要改变,
即至少改变$x$个位置, 若 $x>K$, 说明无解 .

于是若存在答案, 则 $x<=K$,
$x$ 个位置的数要改变, 方案数为
$$\lfloor \frac{M}{d}{\rfloor }^x\text{\hspace{1em}(1)}$$
剩下还需改变 $K-x$ 个位置上的数, 方案数为
$$\left(\genfrac{}{}{0ex}{}{N}{N-x}\right)\ast \lfloor \frac{M}{d}-1{\rfloor }^{K-x}\text{\hspace{1em}(2)}$$

$$\therefore tmp[d]=\lfloor \frac{M}{d}{\rfloor }^x\left(\genfrac{}{}{0ex}{}{N}{N-x}\right)\lfloor \frac{M}{d}-1{\rfloor }^{K-x}\text{\hspace{1em}( (1) * (2) )}$$

由于 $gcd$ 为 $d$, 所以要减去 $gcd$ 是 $Ans[d]$ 倍数的方案
$$Ans[d]=tmp[d]-tmp[2d]-tmp[3d]-...-tmp[nd]$$

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$\mathcal{C}odewithbug$

#include<bits/stdc++.h>
#define reg register

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ c = getchar(), flag = -1; break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 300005;
const int mod = 1000000007;

int N;
int M;
int K;
int x;
int A[maxn];
int jc[maxn];
int rev[maxn];
int tmp[maxn];
int Ans[maxn];

int C(int n, int m){
        int fz = jc[n];
        int fm = 1ll*jc[n-m]*jc[m] % mod;
        fz = 1ll*fz*rev[fm] % mod;
        return fz;
}

int KSM(int a, int b){
        int s = 1;
        while(b){
                if(b & 1) s = 1ll*s*a % mod;
                a = 1ll*a*a % mod;
                b >>= 1;
        }
        return s;
}

void Init(){
        rev[1] = 1;
        for(reg int i = 2; i < maxn; i ++) rev[i] = ((1ll*(-mod/i) * 1ll*rev[mod%i]) %mod + mod) % mod;
        jc[0] = 1;
        for(reg int i = 1; i < maxn; i ++) jc[i] = 1ll*jc[i-1]*i % mod;
}

int main(){
        freopen("gcd.in", "r", stdin);
        freopen("gcd.out", "w", stdout);
        Init();
        N = read(); M = read(); K = read();
        for(reg int i = 1; i <= N; i ++) A[i] = read();
        for(reg int d = 1; d <= M; d ++){
                x = 0;
                for(reg int i = 1; i <= N; i ++) x += ((A[i]%d) != 0);
                if(x > K){ Ans[d] = 0; continue ; }
                int tmp_1 = KSM(M/d, x), tmp_2 = C(N, N-x), tmp_3 = KSM(M/d-1, K-x);
                tmp[d] = (1ll*tmp_1*tmp_2%mod)*1ll*tmp_3 % mod;
        }
        for(reg int d = 1; d <= M; d ++){
                Ans[d] = tmp[d];
                for(reg int i = 2; i*d <= M; i ++) Ans[d] = ((Ans[d] - tmp[i*d])%mod + mod) % mod;
                printf("%d\n", Ans[d]);
        }
        return 0;
}