spoj 2319 BIGSEQ

You are given the sequence of all K-digit binary numbers: 0, 1,..., 2K-1. You need to fully partition the sequence into M chunks. Each chunk must be a consecutive subsequence of the original sequence. Let Si (1 ≤ i ≤ M) be the total number of 1's in all numbers in the ith chunk when written in binary, and let S be the maximum of all Si, i.e. the maximum number of 1's in any chunk. Your goal is to minimize S.

Input

In the first line of input, two numbers, K and M (1 ≤ K ≤ 100, 1 ≤ M ≤ 100, M ≤ 2^K), are given, separated by a single space character.

Output

In one line of the output, write the minimum S that can be obtained by some split. Write it without leading zeros. The result is not guaranteed to fit in a 64-bit integer.

Example

Input:

3 4

Output:

4

题解：

from 算法合集之《浅谈数位类统计问题》

code：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
char ch;
bool ok;
void read(int &x){
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    if (ok) x=-x;
}
const int maxn=105;
int k,n;
const int MAXN=10;
const int maxnum=10000;
struct bignum{
    int len,v[MAXN];
    bignum(){memset(v,0,sizeof(v)),len=1;}
    bignum operator=(const char* num){
        memset(v,0,sizeof(v));
        len=((strlen(num)-1)>>2)+1;
        int j=1,k=0;
        for (int i=strlen(num)-1;i>=0;i--){
            if (j==maxnum) j=1,k++;
            v[k]+=(num[i]-'0')*j;
            j*=10;
        }
        return *this;
    }
    bignum operator=(const int num){
        char a[MAXN<<2];
        sprintf(a,"%d",num);
        *this=a;
        return *this;
    }
    bignum (int num){*this=num;}
    bignum (const char* num){*this=num;}
    bignum operator+(const bignum &a){
        bignum c;
        c.len=max(len,a.len);
        for (int i=0;i<c.len;i++){
            c.v[i]+=v[i]+a.v[i];
            if (c.v[i]>=maxnum) c.v[i+1]+=(c.v[i]/maxnum),c.v[i]%=maxnum;    
        }
        while (c.v[c.len]) c.len++;
        return c;
    }
    bignum operator-(const bignum b){
        bignum a,c;
        a=*this;
        c.len=len;
        for (int i=0;i<len;i++){
            if (a.v[i]<b.v[i]) a.v[i+1]--,a.v[i]+=maxnum;
            c.v[i]=a.v[i]-b.v[i];
        }    
        while (c.len>1&&!(c.v[c.len-1])) c.len--;
        return c;
    }
    bignum operator*(const bignum &a){
        bignum c;
        c.len=len+a.len;
        for (int i=0;i<len;i++)
            for (int j=0;j<a.len;j++){
                c.v[i+j]+=v[i]*a.v[j];
                if (c.v[i+j]>=maxnum) c.v[i+j+1]+=(c.v[i+j]/maxnum),c.v[i+j]%=maxnum;    
            }
        while (c.len>1&&!(c.v[c.len-1])) c.len--;
        return c;
    }
    bignum operator*(const int &a){
        bignum c=a;
        return *this*c;    
    }
    bignum operator/(const int &b){
        bignum c;
        int x=0;
        for (int i=len-1;i>=0;i--){
            c.v[i]=(x*maxnum+v[i])/b;
            x=(x*maxnum+v[i])%b;
        }
        c.len=len;
        while (c.len>1&&!(c.v[c.len-1])) c.len--;
        return c;
    }
    bignum operator+=(const bignum &a){*this=*this+a;return *this;}
    bignum operator-=(const bignum &a){*this=*this-a;return *this;}
    bignum operator*=(const bignum &a){*this=*this*a;return *this;}
    bignum operator/=(const int &a){*this=*this/a;return *this;}
    bool operator < (const bignum &x)const{
        if (len!=x.len) return len<x.len;
        for (int i=len-1;i>=0;i--)
            if (v[i]!=x.v[i]) return v[i]<x.v[i];
        return false;
    }
    bool operator > (const bignum &x)const{return x<*this;}
    bool operator <=(const bignum &x)const{return !(x<*this);}
    bool operator >=(const bignum &x)const{return !(*this<x);}
    bool operator ==(const bignum &x)const{return !(x<*this||*this<x);}
    bool operator !=(const bignum &x)const{return x<*this||*this<x;}
};    
void write(bignum x){
    printf("%d",x.v[x.len-1]);
    for (int i=x.len-2;i>=0;i--) printf("%0*d",4,x.v[i]);
    puts("");
}
void read(bignum &x){
    static char num[MAXN<<2];
    scanf("%s",num);
    x=num;
}
bignum l,r,m,pw2[maxn],sum[maxn],tmp,res,t,xx;
int a[maxn];
void init(){
    pw2[0]=1;
    for (int i=1;i<=101;i++) pw2[i]=pw2[i-1]*2;
    sum[0]=0;
    for (int i=1;i<=101;i++) sum[i]=sum[i-1]*2+pw2[i-1];
}
void calc(){
    tmp=1,res=0;
    for (int i=1;i<=k;i++) if (a[i]) res+=tmp+sum[i-1],tmp+=pw2[i-1];
}
void find(bignum lim){
    bool flag=(lim>=sum[k]);
    int tmp=0;
    for (int i=k;i>=1;i--){
        xx=sum[i-1]+pw2[i-1]*tmp;
        if (lim>=xx) lim=lim-xx,tmp++,a[i]=1;
        else a[i]=0;
    }
    if (!flag){
        for (int i=1;i<=k;i++){
            a[i]^=1;
            if (!a[i]) break;
        }
    }
}
bool check(){
    for (int i=1;i<=k;i++) if (!a[i]) return false;
    return true;
}
bool check(bignum lim){
    bignum t=0;
    int cnt=n;
    memset(a,0,sizeof(a));
    while (cnt--){
        find(t+lim),calc(),t=res;
        if (check()) return true;
    }
    return false;
}
int main(){
    init();
    read(k),read(n);
    l=1,r=sum[k];
    while (l!=r){
        m=(l+r)/2;
        if (check(m)) r=m; else l=m+1;
    }
    write(l);
    return 0;
}