Sum - 走看看

Sum

时间限制: 1 Sec 内存限制: 128 MB

题目描述

求有多少 n位十进制数是 p的倍数且每位之和小于等于 mi(mi=0,1,2,…,m−1,m)，允许前导 0，答案对 998244353 取模。

输入

一行三个整数 n,p,m。(1≤n≤109,1≤p≤50.1≤m≤1000)

输出

输出一行 m+1个正整数，分别表示 mi=0,1,2,…,m−1,m时的答案。

样例输入

2 3 3

样例输出

1 1 1 5
分析：考虑dp，dp[i][(j*10+k)%p][x+k]+=dp[i-1][j][x];
　　　然后考虑倍增；
 $F [x * 2] [(j * 10^{y} + l) % p] [a + b] = F [x] [j] [a] * F [x] [l] [b]$

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <cassert>
#include <ctime>
#define rep(i,m,n) for(i=m;i<=(int)n;i++)
#define inf 0x3f3f3f3f
#define mod 998244353
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
#define ls (rt<<1)
#define rs (rt<<1|1)
#define all(x) x.begin(),x.end()
const int maxn=3e3+10;
const int N=5e4+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qmul(ll p,ll q,ll mo){ll f=0;while(q){if(q&1)f=(f+p)%mo;p=(p+p)%mo;q>>=1;}return f;}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,p;
const int G=3;
int tmp[maxn],wn[maxn],ret[51][maxn],mid[51][maxn],q[51][maxn];
void upd(int &x,int y){x=(0LL+x+y)%mod;}
void getwn(int len)
{
    wn[0]=1;
    wn[1]=qpow(G,(mod-1)/len);
    int i;
    rep(i,2,len) wn[i]=(ll)wn[i-1]*wn[1]%mod;
}

void NTT(int *a,int len,int on){
    int i;
    ll ni=qpow(len,mod-2);
    rep(i,0,len-1){
        int p=0;
        double ce=log(len)/log(2);
        for(int j=0,tp=i;j<ce;j++,tp/=2) p=(p<<1)+(tp%2);
        tmp[p]=a[i];
    }
    for (int m=2;m<=len;m*=2){
        int half=m/2,bei=len/m;
        rep(i,0,half-1){
            ll wi=on>0?wn[i*bei]:wn[len-i*bei];
            for(int j=i;j<len;j+=m){
                ll u=tmp[j],v=(ll)tmp[j+half]*wi%mod;
                tmp[j]=(u+v)%mod;
                tmp[j+half]=(u-v+mod)%mod;
            }
        }
    }
    if (on==-1)
        rep(i,0,len-1) tmp[i]=tmp[i]*ni%mod;
    rep(i,0,len-1) a[i]=tmp[i];
}
int main(){
    int i,j;
    scanf("%d%d%d",&n,&p,&m);
    int len=1;
    m++;
    while(len<m*2)len<<=1;
    m--;
    getwn(len);
    ret[0][0]=1;
    int up=10%p;
    for(i=min(9,m);~i;i--)q[i%p][i]++;
    for(;n;n>>=1)
    {
        if(n&1)
        {
            memset(mid,0,sizeof(mid));
            rep(i,0,p-1)NTT(ret[i],len,1),NTT(q[i],len,1);
            rep(i,0,p-1)rep(j,0,p-1)rep(k,0,len-1)
            {
                upd(mid[(i*up+j)%p][k],(ll)ret[i][k]*q[j][k]%mod);
            }
            rep(i,0,p-1)NTT(mid[i],len,-1),NTT(q[i],len,-1);
            memcpy(ret,mid,sizeof(ret));
            rep(i,0,p-1)rep(j,m+1,len-1)ret[i][j]=0;
        }
        memset(mid,0,sizeof(mid));
        rep(i,0,p-1)NTT(q[i],len,1);
        rep(i,0,p-1)rep(j,0,p-1)rep(k,0,len-1)
        {
            upd(mid[(i*up+j)%p][k],(ll)q[i][k]*q[j][k]%mod);
        }
        rep(i,0,p-1)NTT(mid[i],len,-1);
        memcpy(q,mid,sizeof(q));
        rep(i,0,p-1)rep(j,m+1,len-1)q[i][j]=0;
        up=(ll)up*up%p;
    }
    rep(i,0,m)
    {
        printf("%d%c",ret[0][i],i==m?'
':' ');
        if(i+1<=m)upd(ret[0][i+1],ret[0][i]);
    }
    return 0;
}