spoj 2319 BIGSEQ - Sequence

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：

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<algorithm>
  6 using namespace std;
  7 char ch;
  8 bool ok;
  9 void read(int &x){
 10     for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
 11     for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
 12     if (ok) x=-x;
 13 }
 14 const int maxn=105;
 15 int k,n;
 16 const int MAXN=10;
 17 const int maxnum=10000;
 18 struct bignum{
 19     int len,v[MAXN];
 20     bignum(){memset(v,0,sizeof(v)),len=1;}
 21     bignum operator=(const char* num){
 22         memset(v,0,sizeof(v));
 23         len=((strlen(num)-1)>>2)+1;
 24         int j=1,k=0;
 25         for (int i=strlen(num)-1;i>=0;i--){
 26             if (j==maxnum) j=1,k++;
 27             v[k]+=(num[i]-'0')*j;
 28             j*=10;
 29         }
 30         return *this;
 31     }
 32     bignum operator=(const int num){
 33         char a[MAXN<<2];
 34         sprintf(a,"%d",num);
 35         *this=a;
 36         return *this;
 37     }
 38     bignum (int num){*this=num;}
 39     bignum (const char* num){*this=num;}
 40     bignum operator+(const bignum &a){
 41         bignum c;
 42         c.len=max(len,a.len);
 43         for (int i=0;i<c.len;i++){
 44             c.v[i]+=v[i]+a.v[i];
 45             if (c.v[i]>=maxnum) c.v[i+1]+=(c.v[i]/maxnum),c.v[i]%=maxnum;    
 46         }
 47         while (c.v[c.len]) c.len++;
 48         return c;
 49     }
 50     bignum operator-(const bignum b){
 51         bignum a,c;
 52         a=*this;
 53         c.len=len;
 54         for (int i=0;i<len;i++){
 55             if (a.v[i]<b.v[i]) a.v[i+1]--,a.v[i]+=maxnum;
 56             c.v[i]=a.v[i]-b.v[i];
 57         }    
 58         while (c.len>1&&!(c.v[c.len-1])) c.len--;
 59         return c;
 60     }
 61     bignum operator*(const bignum &a){
 62         bignum c;
 63         c.len=len+a.len;
 64         for (int i=0;i<len;i++)
 65             for (int j=0;j<a.len;j++){
 66                 c.v[i+j]+=v[i]*a.v[j];
 67                 if (c.v[i+j]>=maxnum) c.v[i+j+1]+=(c.v[i+j]/maxnum),c.v[i+j]%=maxnum;    
 68             }
 69         while (c.len>1&&!(c.v[c.len-1])) c.len--;
 70         return c;
 71     }
 72     bignum operator*(const int &a){
 73         bignum c=a;
 74         return *this*c;    
 75     }
 76     bignum operator/(const int &b){
 77         bignum c;
 78         int x=0;
 79         for (int i=len-1;i>=0;i--){
 80             c.v[i]=(x*maxnum+v[i])/b;
 81             x=(x*maxnum+v[i])%b;
 82         }
 83         c.len=len;
 84         while (c.len>1&&!(c.v[c.len-1])) c.len--;
 85         return c;
 86     }
 87     bignum operator+=(const bignum &a){*this=*this+a;return *this;}
 88     bignum operator-=(const bignum &a){*this=*this-a;return *this;}
 89     bignum operator*=(const bignum &a){*this=*this*a;return *this;}
 90     bignum operator/=(const int &a){*this=*this/a;return *this;}
 91     bool operator < (const bignum &x)const{
 92         if (len!=x.len) return len<x.len;
 93         for (int i=len-1;i>=0;i--)
 94             if (v[i]!=x.v[i]) return v[i]<x.v[i];
 95         return false;
 96     }
 97     bool operator > (const bignum &x)const{return x<*this;}
 98     bool operator <=(const bignum &x)const{return !(x<*this);}
 99     bool operator >=(const bignum &x)const{return !(*this<x);}
100     bool operator ==(const bignum &x)const{return !(x<*this||*this<x);}
101     bool operator !=(const bignum &x)const{return x<*this||*this<x;}
102 };    
103 void write(bignum x){
104     printf("%d",x.v[x.len-1]);
105     for (int i=x.len-2;i>=0;i--) printf("%0*d",4,x.v[i]);
106     puts("");
107 }
108 void read(bignum &x){
109     static char num[MAXN<<2];
110     scanf("%s",num);
111     x=num;
112 }
113 bignum l,r,m,pw2[maxn],sum[maxn],tmp,res,t,xx;
114 int a[maxn];
115 void init(){
116     pw2[0]=1;
117     for (int i=1;i<=101;i++) pw2[i]=pw2[i-1]*2;
118     sum[0]=0;
119     for (int i=1;i<=101;i++) sum[i]=sum[i-1]*2+pw2[i-1];
120 }
121 void calc(){
122     tmp=1,res=0;
123     for (int i=1;i<=k;i++) if (a[i]) res+=tmp+sum[i-1],tmp+=pw2[i-1];
124 }
125 void find(bignum lim){
126     bool flag=(lim>=sum[k]);
127     int tmp=0;
128     for (int i=k;i>=1;i--){
129         xx=sum[i-1]+pw2[i-1]*tmp;
130         if (lim>=xx) lim=lim-xx,tmp++,a[i]=1;
131         else a[i]=0;
132     }
133     if (!flag){
134         for (int i=1;i<=k;i++){
135             a[i]^=1;
136             if (!a[i]) break;
137         }
138     }
139 }
140 bool check(){
141     for (int i=1;i<=k;i++) if (!a[i]) return false;
142     return true;
143 }
144 bool check(bignum lim){
145     bignum t=0;
146     int cnt=n;
147     memset(a,0,sizeof(a));
148     while (cnt--){
149         find(t+lim),calc(),t=res;
150         if (check()) return true;
151     }
152     return false;
153 }
154 int main(){
155     init();
156     read(k),read(n);
157     l=1,r=sum[k];
158     while (l!=r){
159         m=(l+r)/2;
160         if (check(m)) r=m; else l=m+1;
161     }
162     write(l);
163     return 0;
164 }

 

转载于:https://www.cnblogs.com/chenyushuo/p/5239840.html