BZOJ4737 组合数问题（卢卡斯定理+数位dp）

　　不妨不管j<=i的限制。由卢卡斯定理，C(i,j) mod k=0相当于k进制下存在某位上j大于i。容易想到数位dp，即设f[x][0/1][0/1][0/1]为到第x位时是否有某位上j>i，是否卡n、m的限制的方案数。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100
#define P 1000000007
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
ll n,m;
int T,k,ans,f[N][][][],a[N],b[N];
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int calc(ll n,ll m)
{
    int t=-;
    while (n) a[++t]=n%k,n/=k;
    for (int i=;i<=t;i++) b[i]=m%k,m/=k;
    memset(f,,sizeof(f));
    f[t+][][][]=;
    for (int i=t;~i;i--)
        for (int j=;j<=;j++)
            for (int x=;x<=;x++)
                for (int y=;y<=;y++)
                    for (int p=x;p<=;p++)
                        for (int q=y;q<=;q++)
                        {
                            int ln=x==?a[i]:,rn=x==?a[i]:(p==?a[i]-:k-),lm=y==?b[i]:,rm=y==?b[i]:(q==?b[i]-:k-);
                            int s=;
                            for (int u=ln;u<=rn;u++)
                                for (int v=lm;v<=rm;v++)
                                if (v<=u) s++;
                            inc(f[i][j][x][y],1ll*f[i+][j][p][q]*s%P);
                            if (j) inc(f[i][j][x][y],1ll*f[i+][j-][p][q]*((rn-ln+)*(rm-lm+)-s)%P),
                            inc(f[i][j][x][y],1ll*f[i+][j][p][q]*((rn-ln+)*(rm-lm+)-s)%P);
                        }
    return ((f[][][][]+f[][][][])%P+(f[][][][]+f[][][][])%P)%P;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4737.in","r",stdin);
    freopen("bzoj4737.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read(),k=read();
    while (T--)
    {
        cin>>n>>m;m=min(n,m);
        if (m&) ans=(P-(m%P)*((m+>>)%P)%P)%P;
        else ans=(P-((m+)%P)*((m>>)%P)%P)%P;
        inc(ans,calc(n,m));
        cout<<ans<<endl;
    }
    return ;
}