Codeforces 9C Hexadecimal's Numbers （解法汇总）
题目链接：http://codeforces.com/problemset/problem/9/C
题面：
C. Hexadecimal's Numbers
One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.
But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.
Input data contains the only number n (1 ≤ n ≤ 109).
Output the only number — answer to the problem.
10
2
For n = 10 the answer includes numbers 1 and 10.
题目大意：
问1-n中，只由01组成的数有多少个。
解题：
在网上看到很多解法，此处罗列一下。
解法一：（函数调用）
来源：http://blog.csdn.net/u013076044/article/details/39032395
代码：
#include <iostream>
using namespace std;
int n;
int ans;
void f(int x){
if(x>n) return ;
ans++;
f(x*10);
f(x*10+1);
}
int main(){
cin>>n;
ans=0;
f(1);
cout<<ans<<endl;
}
解法二：（找规律）
来源：http://blog.csdn.net/lvshubao1314/article/details/28693243
代码：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
char a[10];
int sum[11]={0,1,2,4,8,16,32,64,128,256,512};
int main()
{
while(~scanf("%s",a+1))
{
int n=strlen(a+1);
int sum1=0;
for(int i=1;i<=n;i++)
{
if(a[i]>'1')
{
for(int j=1;j<=n-i+1;j++)
sum1+=sum[j];
break;
}
if(a[i]=='1')
sum1+=sum[n-i+1];
}
printf("%d\n",sum1);
}
return 0;
}
解法三：（搜索）
来源：蠢蠢的我的解法。
代码：
#include <iostream>
#include <cmath>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
vector <int> store;
void add(string s)
{
int tmp=0,t=1;
for(int i=s.length()-1;i>=0;i--)
{
tmp+=(t*(s[i]-'0'));
t*=10;
}
store.push_back(tmp);
}
void dfs(string s,int x,int len)
{
if(x)
{
s+='1';
add(s);
if(len<9)
{
dfs(s,1,len+1);
dfs(s,0,len+1);
}
else return;
}
else
{
s+='0';
add(s);
if(len<9)
{
dfs(s,1,len+1);
dfs(s,0,len+1);
}
else return;
}
}
int main()
{
dfs("1",0,1);
dfs("1",1,1);
store.push_back(1);
sort(store.begin(),store.end());
int n,ans=0,sz=store.size();
cin>>n;
for(int i=0;i<sz;i++)
{
if(store[i]<=n)
ans++;
else break;
}
cout<<ans<<endl;
return 0;
}
解法四：（数位dp）
所谓数位dp实际与找规律相似。
来源：扒队友的
代码：
#include <cstdio>
#include <iostream>
using namespace std;
int dp[15];
int main()
{
int n,dig[15],i,len,ans;
dp[0]=1;
dp[1]=2;
for(int i=2;i<=10;i++)
dp[i]=dp[i-1]*2;
scanf("%d",&n);
i=1;
while(n)
{
dig[i++]=n%10;
n=n/10;
}
len=i-1;
if(dig[len]==1)
ans=dp[len-1]-1;
else
{
ans=dp[len-1]-1+dp[len-1];
printf("%d\n",ans);
return 0;
}
for(i=len-1;i>=1;i--)
{
if(dig[i]>1)
{
ans+=dp[i-1]*2;
break;
}
else if(dig[i]==1)
ans+=dp[i-1];
}
if(i==0)
ans++;
printf("%d\n",ans);
return 0;
}