Codeforces Round #247(Div. 2) D. Random Task 二分+前缀和

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.

Examples

input

1 1

output

1

input

3 2

output

5

题意：

给的m和k，找到一个n，使得区间[n+1,2*n]里的数中，恰好有m个数满足要求：该数的二进制表示里恰好有k个1.

题解：

在k相同的情况下，n单调，可以二分。

重点是求出某个区间满足要求的数的个数。

我们设cal(x)为区间[0,x]中，二进制表示中1个个数为k的数的个数，那么对于一个二分的答案Mid，对应的[Mid+1,2*Mid]满足要求的数的个数为cal(2*Mid)-cal(Mid).

例如二进制x=101101 k=3

从高位到底为枚举每一位i：第一位为1，将其置0后，后面可以有任意k个1，故res+=Com(i,k-cnt)，Com为组合数，cnt为该位前面的1的个数。第2位为0，跳过；第3位为1，

将该置0，则前面有2个1被置0，则后面还要添加k-2个1。res+=Com(i,k-2)，依次计算。

/****************
*PID:431d div2
*Auth:Jonariguez
*****************
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

const int maxn=100000+10;
LL m,k;

LL Com(LL n,LL m){
if(m==0) return 1LL;
LL res=n--,i;
for(i=2;i<=m;i++){
res*=n;res/=i;
n--;
}
return res;
}

LL cal(LL n){
LL res=0,i,cnt=0;
for(i=62;i>=0;i--){
if((n>>i)&1){
if(k-cnt>=0)
res+=Com(i,k-cnt);
cnt++;
}
}
return res;
}

int main()
{
int i,j;
while(scanf("%I64d%I64d",&m,&k)!=EOF){
LL l=1,r=1e18,Mid,res=-1;
while(l<r-1){
Mid=l+(r-l)/2;
LL temp=cal(2*Mid)-cal(Mid);
if(temp==m){
res=Mid;break;
}else if(temp<m)
l=Mid+1;
else r=Mid;
}
// printf("cal=%I64d\n",cal(2));
if(res>=0)
printf("%I64d\n",res);
else printf("%I64d\n",l);
}
return 0;
}