bzoj 1192

http://www.lydsy.com/JudgeOnline/problem.php?id=1192

好像学过一个东西：

[0..2^(N+1)-1]内的数都的都可以由2^0,2^1,...,2^N这N+1个数中若干个相加得到。

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<fstream>

#include<algorithm>

#include<cstring>

#include<string>

#include<cmath>

#include<queue>

#include<stack>

#include<map>

#include<utility>

#include<set>

#include<bitset>

#include<vector>

#include<functional>

#include<deque>

#include<cctype>

#include<climits>

#include<complex>

//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj

using namespace std;

typedef long long LL;

typedef double DB;

typedef pair<int,int> PII;

typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))

#define mmcy(a,b) memcpy(a,b,sizeof(a))

#define fill(a,l,r,v) fill(a+l,a+r+1,v)

#define re(i,a,b)  for(i=(a);i<=(b);i++)

#define red(i,a,b) for(i=(a);i>=(b);i--)

#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)

#define fi first

#define se second

#define m_p(a,b) make_pair(a,b)

#define SF scanf

#define PF printf

#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}

template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}

template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-;

inline int sgn(DB x){if(abs(x)<EPS)return ;return(x>)?:-;}

const DB Pi=acos(-1.0);

inline int gint()

  {

        int res=;bool neg=;char z;

        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());

        if(z==EOF)return ;

        if(z=='-'){neg=;z=getchar();}

        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());

        return (neg)?-res:res;

    }

inline LL gll()

  {

      LL res=;bool neg=;char z;

        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());

        if(z==EOF)return ;

        if(z=='-'){neg=;z=getchar();}

        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());

        return (neg)?-res:res;

    }

int m,cnt,p;

int main()

  {

      freopen("bzoj1192.in","r",stdin);

        freopen("bzoj1192.out","w",stdout);

        m=gint();

        cnt=;p=;

        while(p-<m)p<<=,cnt++;

        PF("%d\n",cnt+);

        return ;

    }