Codeforces 431 D. Random Task

很巧妙的单调性......

n,n+1,n+2,.....2*n-2

   n+1,n+2,.....2*n-2,2*n-1,2*n

中间一段是相同的,n和2*n里的1是一样多的所以只有2*n-1不一样.....这是满足单调性的(monotone)

然后就是数位DP了.....

D. Random Task

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day, after a difficult lecture a diligent student Sasha saw a graffitied desk in the classroom. She came closer and read: "Find such positive integer n, that among numbers n + 1, n + 2, ..., 2·n there are exactly m numbers which binary representation contains exactlyk digits one".

The girl got interested in the task and she asked you to help her solve it. Sasha knows that you are afraid of large numbers, so she guaranteed that there is an answer that doesn't exceed 1018.

Input

The first line contains two space-separated integers, m and k (0 ≤ m ≤ 1018; 1 ≤ k ≤ 64).

Output

Print the required number n (1 ≤ n ≤ 1018). If there are multiple answers, print any of them.

Sample test(s)

input

1 1

output

1

input

3 2

output

5

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

LL m,k;
LL C[70][70];

int init()
{
    for(int i=0;i<70;i++)
        C[i][0]=C[i][i]=1LL;
    for(int i=2;i<70;i++)
    {
        for(int j=1;j<i;j++)
        {
            C[i][j]=C[i-1][j-1]+C[i-1][j];
        }
    }
}

LL get_sum_one(LL x)
{
    LL ret=0;
    int cur=0;
    for(int i=63;i>=0;i--)
    {
        if(x&(1LL<<i))
        {
            if(k-cur>=0) ret+=C[i][k-cur];
            cur++;
        }
    }
    if(cur==k) ret++;
    return ret;
}

LL ck(LL x)
{
    LL sum0=get_sum_one(2*x);
    LL sum1=get_sum_one(x);
    return sum0-sum1;
}

int main()
{
    init();
    LL x;
    cin>>m>>k;
    LL low=1LL,high=1000000000000000000LL,mid,ans=-1;
    while(low<=high)
    {
        mid=(low+high)/2LL;
        LL t=ck(mid);
        if(t>=m) high=mid-1;
        else low=mid+1;
        if(t==m)
        {
            ans=mid;
            break;
        }
    }
    cout<<ans<<endl;
    return 0;
}