题目大意
给定一个集合 ({1,2,ldots,n})，要求你从中选出 (m) 个数，且这 (m) 个数的和是 (k)。问方案数 (mod 998244353)
(0leq k<n<998244353,mleq n)
题解
先不考虑选的数的个数的限制。
显然答案的 OGF 为
[F(x)=prod_{i=0}^{n-1}(1+x^i)
]
考虑答案的式子
[egin{align}
ans&=frac{1}{n}sum_{i=0}^{n-1}F(omega_n^i)omega_n^{-ik}\
&=frac{1}{n}sum_{i=0}^{n-1}prod_{j=0}^{n-1}(1+omega_n^{ij})omega_n^{-ik}
end{align}
]
记 (d=gcd(n,i))，那么
[egin{align}
ans&=frac{1}{n}sum_{i=0}^{n-1}prod_{j=0}^{n-1}(1+omega_n^{ij})omega_n^{-ik}\
&=frac{1}{n}sum_{i=0}^{n-1}prod_{j=0}^{frac{n}{d}-1}{(1+omega_frac{n}{d}^{ij})}^domega_n^{-ik}\
end{align}
]
容易发现，(prod_{i=0}^{n-1}(1+omega_n^i)=1-{(-1)}^n)
[egin{align}
ans&=frac{1}{n}sum_{i=0}^{n-1}prod_{j=0}^{frac{n}{d}-1}{(1+omega_frac{n}{d}^{ij})}^domega_n^{-ik}\
&=frac{1}{n}sum_{i=0}^{n-1}{(prod_{j=0}^{frac{n}{d}-1}{(1+omega_frac{n}{d}^{ij})})}^domega_n^{-ik}\
&=frac{1}{n}sum_{i=0}^{n-1}{(1-{(-1)}^frac{n}{d})}^domega_n^{-ik}\
&=frac{1}{n}sum_{dmid n}{(1-{(-1)}^frac{n}{d})}^d(sum_{gcd(i,n)=d}omega_n^{-ik})\
&=frac{1}{n}sum_{dmid n}{(1-{(-1)}^frac{n}{d})}^d(sum_{gcd(i,frac{n}{d})=1}omega_frac{n}{d}^{-ik})\
&=frac{1}{n}sum_{dmid n}{(1-{(-1)}^frac{n}{d})}^d(sum_{jmid frac{n}{d}}mu(j)sum_{i=0}^{frac{n}{dj}-1}omega_frac{n}{dj}^{-ik})\
&=frac{1}{n}sum_{dmid n}{(1-{(-1)}^frac{n}{d})}^d(sum_{jmid frac{n}{d}}mu(j)[frac{n}{dj}mid k]frac{n}{dj})\
end{align}
]
现在要加上 (m) 这个限制，只需要多加一个元 (y)，把初始的 OGF 改为
[prod_{i=0}^{n-1}(1+x^iy)
]
推导过程中把 (y) 当做一个常量，就像 (x) 一样。最后的式子为
[[y^m]frac{1}{n}sum_{dmid n}{(1-{(-y)}^frac{n}{d})}^d(sum_{jmid frac{n}{d}}mu(j)[frac{n}{dj}mid k]frac{n}{dj})\
]
还有一个问题就是要在二项式展开的时候计算组合数。直接分段打表就好了。
时间复杂度：(O(sigma_0(n)^2))。
代码
阶乘的表删掉了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<assert.h>
using namespace std;
typedef long long ll;
const ll p=998244353;
ll fp(ll a,ll b)
{
ll s=1;
for(;b;b>>=1,a=a*a%p)
if(b&1)
s=s*a%p;
return s;
}
int n,m,k;
ll getmiu(int x)
{
ll s=1;
for(int i=2;i*i<=x;i++)
if(x%i==0)
{
int tmp=0;
while(x%i==0)
{
tmp++;
x/=i;
}
if(tmp>=2)
return 0;
s=-s;
}
if(x!=1)
s=-s;
return s;
}
int c[1];
const int N=10000000;
const int D=200000;
int fac[N+10];
ll factorial(int n)
{
if(n<=N)
return fac[n];
ll s=c[n/D];
for(int i=n/D*D+1;i<=n;i++)
s=s*i%p;
return s;
}
ll binom(int x,int y)
{
return x>=y&&y>=0?factorial(x)*fp(factorial(y)*factorial(x-y)%p,p-2)%p:0;
}
void init()
{
fac[0]=1;
for(int i=1;i<=N;i++)
fac[i]=(ll)fac[i-1]*i%p;
}
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
int exgcd(int a,int b,int &x,int &y)
{
if(!b)
{
x=1;
y=0;
return a;
}
int d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
int get(int a)
{
int x,y;
int d=exgcd(a,p-1,x,y);
if(x<0)
{
printf("%d %d
%lld
",x,y,(ll)x*a+y*(p-1));
x+=(p-1)/d;
y-=a/d;
printf("%d %d
%lld
",x,y,(ll)x*a+y*(p-1));
}
return x;
}
ll mu[10000];
int d[10000];
int t=0;
int query(int x)
{
return mu[lower_bound(d+1,d+t+1,x)-d];
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
#endif
init();
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i*i<=n;i++)
if(n%i==0)
{
d[++t]=i;
if(i*i!=n)
d[++t]=n/i;
}
sort(d+1,d+t+1);
for(int i=1;i<=t;i++)
mu[i]=getmiu(d[i]);
ll ans=0;
for(int i=1;i<=t;i++)
{
if(m%(n/d[i]))
continue;
int w=m/(n/d[i]);
int v=(w&1?-((n/d[i])&1?-1:1):1);
ll s1=binom(d[i],w);
ll s2=0;
for(int j=1;j<=t;j++)
if((n/d[i])%d[j]==0)
if(k%(n/d[i]/d[j])==0)
s2=(s2+mu[j]*n/d[i]/d[j])%p;
ans=(ans+s1*s2*v)%p;
}
ans=ans*fp(n,p-2)%p;
ans=(ans+p)%p;
printf("%lld
",ans);
return 0;
}