Number theory difference + Lagrange interpolation lgP5430 solution

Posted by konn on Mon, 10 Jan 2022 10:07:50 +0100

New \ (O(k+\log n) \) approach.

Consider calculating the contribution of each monkey to the answer.

Make a table:

1 1 2 4 8 16 32 ...

It can be seen that the contribution of the $I$monkey to the answer is \ (i^k \times 2^{n-i-1} \), in particular, the contribution of the last monkey to the answer is \ (n^k \).

Written as persimmon:

$n^k+\sum_{i=1}^{n-1}i^k \times 2^{n-i-1}$

$n^k+2^{n-1} \times (\sum_{i=1}^{n-1}i^k \times (2^{-1})^i)$

We just need to calculate \ (\ sum {I = 1} ^ {n-1} I ^ k \ times (2 ^ {- 1}) ^ I \).

Then we found that the persimmon was CODECHEF qpolysum, and then we finished it.

Let's write it down

qpolysum is different from this question, that is \ (i \) starts from \ (0 \), but it doesn't make any difference, because what you want to lose is a \ (0 \)(

$S(n)=\sum_{i=0}^{n-1}i^k \times m^i$

In this question, it is equivalent to \ (m=2^{-1} \).

However, this practice guessed a very strange conclusion, and the practice came from the school OJ discussion area(

We guess that \ (s (n) = m ^ n (g (n) - G (0) \), where \ (G(x) \) is a polynomial of no more than \ (k \).

Proof can see This blog I won't tell you I can't understand

Then make a difference:

$S(n)-S(n-1)=m^n G(n) - m^{n-1} G(n-1) = (n-1)^k \times m^{n-1}$

$G(n) = \frac {(n-1)^k + G(n-1)} m$

Let \ (G(0)=x \), then we use \ (x \) to represent \ (G(n) \) when \ (n \) is any value.

Because the highest degree of this polynomial is \ (k \), and the difference for a \ (k \) polynomial is \ (0 \) after \ (k+1 \) times, we obtain the following result after \ (G(x) \) difference \ (k+1 \) times:

$\sum_{i=0}^{k+1} (-1)^{i+1} \binom {k+1} i G(k+1-i) = 0$

We can use \ (x \) to represent \ (G(0) \sim G(k+1) \), then solve a unary linear equation to obtain \ (x \), and bring in the value of \ (G(1) \sim G(k+1) \).

Now we can use Lagrange interpolation to calculate \ (G(n) \), and the answer is \ (m^nG(n)-G(0) \).

$$i^k$$ can use linear sieve, so the complexity is \ (O(k+\log n) \).

what? You stuck my space\ (10 \) too many arrays???

In fact, it can be reduced to \ (7 \) arrays of \ (\ rm int \) and \ (1 \) arrays of \ (\ rm bool \).

My linear sieve directly records the minimum prime factor rather than whether it is a prime number, which can be changed to the latter.

Then, when using the \ (q \) and \ (p \) arrays, the \ (x \) and \ (y \) arrays will no longer be used, so you can directly use \ (x \) and \ (y \) instead of \ (q \) and \ (p \).

code:

#include<cstdio>
#include<cctype>
const int M=2e7+5,mod=1e9+7;
int k,n1,n2,top,x[M],y[M],idk[M],pri[M],fac[M],ifac[M];bool zhi[M];
int G[M];
char s;long long t1,t2;
while(isdigit(s=getchar())){
t1=n1*10ll+(s^48);n1=t1>=mod?t1%mod:t1;
t2=n2*10ll+(s^48);n2=t2>=mod-1?t2%(mod-1):t2;
}
}
return a+b>=mod?a+b-mod:a+b;
}
inline int Del(const int&a,const int&b){
return b>a?a-b+mod:a-b;
}
inline int C(const int&n,const int&m){
return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}
inline int pow(int a,int b){
int ans=1;
for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)ans=1ll*ans*a%mod;
return ans;
}
inline void sieve(const int&M){
register int i,j,x;idk=1;
for(i=2;i<=M;++i){
if(!zhi[i])pri[++top]=i,idk[i]=pow(i,k);
for(j=1;j<=top&&(x=i*pri[j])<=M;++j){
idk[x]=1ll*idk[i]*idk[pri[j]]%mod;zhi[x]=true;
if(!(i%pri[j]))break;
}
}
}
inline int Inter(const int&n){
register int i,tmp,ans=0;
x=y[k+2]=1;
for(i=1;i<=k+1;++i)x[i]=1ll*x[i-1]*Del(n,i)%mod;
for(i=k+1;i>=1;--i)y[i]=1ll*y[i+1]*Del(n,i)%mod;
for(i=1;i<=k+1;++i){
if(k+1-i&1)ans=Del(ans,1ll*1ll*x[i-1]*y[i+1]%mod*G[i]%mod*ifac[i-1]%mod*ifac[k+1-i]%mod);
}
return ans;
}
signed main(){
register int i,X=0,Y=0;
for(i=2;i<=k+1;++i)fac[i]=1ll*fac[i-1]*i%mod,ifac[i]=1ll*(mod-mod/i)*ifac[mod%i]%mod;
for(i=2;i<=k+1;++i)ifac[i]=1ll*ifac[i-1]*ifac[i]%mod;
for(i=0;i<=k+1;++i){
if(i&1){