題目大意
給定長度爲
答案對1e9+7取模
Data Constraint
題解
稍微轉化一下模型。
每次不枚舉上一個
那最後
其中
假設已經求出了
那麼就等於說要將
時間複雜度:
SRC
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std ;
#define N 100000 + 10
typedef long long ll ;
const int MO = 1e9 + 7 ;
bool flag[N] ;
int Pri[N] , P[N] , Num[N] ;
int fac[2*N] , _fac[2*N] ;
int f[N] , g[N] ;
int T , n , m , x ;
ll Power( ll x , int k ) {
ll s = 1 ;
while ( k ) {
if ( k & 1 ) s = (ll)s * x % MO ;
x = (ll)x * x % MO ;
k /= 2 ;
}
return s ;
}
ll C( int n , int m ) { return (ll)fac[n] * _fac[m] % MO * _fac[n-m] % MO ; }
void Pre() {
fac[0] = _fac[0] = 1 ;
for (int i = 1 ; i < 2 * N ; i ++ ) {
fac[i] = (ll)fac[i-1] * i % MO ;
_fac[i] = Power( fac[i] , MO - 2 ) ;
}
for (int i = 2 ; i <= 100000 ; i ++ ) {
if ( !flag[i] ) {
Pri[++Pri[0]] = i ;
P[i] = i ;
}
for (int j = 1 ; j <= Pri[0] ; j ++ ) {
if ( 1ll * i * Pri[j] > 100000 ) break ;
flag[i*Pri[j]] = 1 ;
P[i*Pri[j]] = Pri[j] ;
if ( i % Pri[j] == 0 ) break ;
}
}
}
void Calc() {
Num[1] = 1 ;
for (int i = 2 ; i <= n ; i ++ ) {
if ( !flag[i] ) { Num[i] = m ; continue ; }
Num[i] = 1 ;
int x = i ;
while ( x > 1 ) {
int tot = 0 ;
int d = P[x] ;
while ( x % d == 0 ) {
tot ++ ;
x /= d ;
}
Num[i] = (ll)Num[i] * C( tot + m - 1 , m - 1 ) % MO ;
}
}
}
int main() {
freopen( "b.in" , "r" , stdin ) ;
freopen( "b.out" , "w" , stdout ) ;
Pre() ;
scanf( "%d" , &T ) ;
while ( T -- ) {
memset( f , 0 , sizeof(f) ) ;
memset( g , 0 , sizeof(g) ) ;
scanf( "%d%d" , &n , &m ) ;
Calc() ;
for (int i = 1 ; i <= n ; i ++ ) scanf( "%d" , &f[i] ) ;
for (int i = 1 ; i <= n ; i ++ ) {
for (int k = 1 ; 1ll * k * i <= n ; k ++ ) {
g[i*k] = (g[i*k] + (ll)f[i] * Num[k] % MO) % MO ;
}
}
for (int i = 1 ; i <= n ; i ++ ) printf( "%d " , g[i] ) ;
printf( "\n" ) ;
}
return 0 ;
}
以上.