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pojusing 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-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);inline int gint(){int res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());return (neg)?-res:res; }
inline LL gll(){LL res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',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=0;p=2;while(p-1<m)p<<=1,cnt++;PF("%d\n",cnt+1);return 0;}