51nod1769(dirichlet convolution)

这个题与上次那个题类似，也是个狄雷克卷积的。
但是这题k很大，不能用快速幂了，那我们就要充分考虑 multiplicative function的性质。具体的过程51nod上面给了很清楚了，主要是给出线性筛部分的代码（这个也是借鉴别人的，侵删）。还有要开读入挂。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const LL mod = 1e9 + 7;
const int maxn = 500000 + 5;
LL inv[100];
LL f[maxn],g[maxn],fg[maxn];
LL n,k;
LL In()
{
    char c = getchar();
    LL num = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) {
        num = num * 10 + c - '0';
        if(num >= mod)
            num %= mod;
        c = getchar();
    }
    return num;
}

LL mod_pow(LL a, LL n){
    LL ret = 1;
    while(n > 0){
        if(n & 1) ret = ret * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ret;
}

void Out(LL x)
{
    if(x > 9)
        Out(x / 10);
    putchar(x % 10 + '0');
}

void init(){
    inv[0] = 1;
    for(int i = 1;i < 100;i++){
        inv[i] = mod_pow(i,mod - 2);
    }
}

void dirichlet(LL f[],LL g[],LL fg[],int n){
    for(int i = 1;i <= n;i++) fg[i] = 0;
    for(int i = 1;i * i <= n;i++){
        for(int j = i;j * i <= n;j++){
            if(i == j) fg[i * j] = (fg[i * j] + f[i] * g[j]) % mod;
            else fg[i * j] = (fg[i * j] + f[i] * g[j] + f[j] * g[i]) % mod;
        }
    }
}

void solve(int k){
    g[1] = 1;
    LL *prime = fg;
    int tot = 0;
    for(int i = 2;i <= n;i++){
        if(!prime[i]) prime[tot++] = i,g[i] = k;
        for(int j = 0;j < tot && prime[j] * i <= n;j++){
            prime[prime[j] * i] = 1;
            int r = 1,x = i;
            while(x % prime[j] == 0) r++,x /= prime[j];
            g[i * prime[j]] = 1LL * g[i] * (r + k - 1) % mod * inv[r] % mod;
            if(r != 1) break;
        }
    }
    dirichlet(f, g, fg, (int)n);
    for(int i = 1;i <= n;i++){
        Out(fg[i]);
        if(i < n) putchar(' ');
        else putchar('\n');
    }
}



int main(int argc, const char * argv[]) {
    init();
    n = In();
    k = In();
    for(int i = 1;i <= n;i++) f[i] = In();
    solve(k);
    return 0;
}