[清华集训2016]组合数问题

[清华集训2016]组合数问题

UOJ
BZOJ
通过Lucas定理把题面转化一下
相当于求i<=n,j<=m在k进制下有多少(i,j)满足i的每一位都大于等于j
数位dp求解

#define ll long long
#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
ll re(){
	ll x=0,w=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*w;
}
int t,k,w1,w2,ans,N[70],M[70],f[70][2][2];
ll n,m;
void add(int&x,int y){x=(x+y)%mod;}
int dfs(int p,int a,int b){
	if(!p)return 1;
	if(~f[p][a][b])return f[p][a][b];
	int g1=a?N[p]:k-1,g2=b?M[p]:k-1,res=0;
	for(int i=0;i<=g1;i++)
		for(int j=0;j<=i&&j<=g2;j++)
			add(res,dfs(p-1,a&&i==g1,b&&j==g2));
	return f[p][a][b]=res;
}
int main(){
    t=re(),k=re();
	while(t--){
		n=re(),m=min(n,re());w1=w2=0;
		ans=1ll*((m+1)%mod*((m+2)%mod)%mod*500000004%mod+(n-m)%mod*((m+1)%mod)%mod)%mod;
		memset(N,0,sizeof(N));
		memset(M,0,sizeof(M));
		while(n){N[++w1]=n%k;n/=k;}
		while(m){M[++w2]=m%k;m/=k;}
		memset(f,-1,sizeof(f));
		add(ans,mod-dfs(w1,1,1));
		printf("%d\n",ans);
	}
	return 0;
}