数位dp 二进制问题

 

 

#include <bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define pb push_back
#define inf 0x3f3f3f3f

ll n;int k;
int a[72],cnt;
ll dp[72][2][72];


void init(){
	dp[1][0][0]=1;
	dp[1][1][1]=1;
	for (int i=2;i<72;i++){


		for (int j=1;j<=i;j++){
			dp[i][1][j]+=dp[i-1][1][j-1];
			dp[i][1][j]+=dp[i-1][0][j-1];
		}
		
		for (int j=0;j<i;j++){
			dp[i][0][j]+=dp[i-1][0][j];
			dp[i][0][j]+=dp[i-1][1][j];
		}
		
	}
	
	
	
}

ll dp1(ll x){
	cnt=0;
	if (!x) return 0;
	while (x) a[++cnt]=x%2, x/=2;
	ll ans=0;int mid=0;
	for (int i=cnt;i>0;i--){
		if (a[i]==1){
			if (mid>k) break;
			ans+=dp[i][0][k-mid];
			mid++;



		}
			
	}

	return ans;
	
}


int main(){
	init();
	cin>>n>>k;
	cout<<dp1(n)<<endl;
	
	
}