Jason and Number

Problem Description

　　Given a number N, Jason will connect from the first prime number to the Nth prime to form a new number.

　　Unfortunately, Jason has gone crazy, and he will repeat operation:

　　1、delete the number in the even position;

　　2、delete the number in the odd position.

　　Now, Jason wants to know what the last number is.

　　

　　For example, if N is equal to 6, the new number is 23571113.

　　After deleting the number in even position, the new number is 2511

　　After deleting the number in odd position, the new number is 51

　　After deleting the number in even position, the new number is 5

　　Therefore, the last number is 5.

Input

　　The input will consist of a series of test cases, and each test case contains an integers N (N <= 20000).

Output

　　For each test case, output what the last number is.

Sample Input

6

Sample Output

5

//解题报告：

在字符串长度为[1,3)时 输出第1个子符；

在字符串长度为[3,11)时 输出第3(1+2^1)个子符；

在字符串长度为[11,23)时
输出第11(3+2^3)个子符；

在字符串长度为[43,171)时
输出第43（11+2^5)个子符；

在字符串长度为[173,683)时
输出第171(43+2^7)个子符;

......;

//标程：

#include<stdio.h>

#include<cmath>

#include<algorithm>

#include<string>

#include<iostream>

#include<string.h>

#define Max 300010

using namespace std;

int prime[Max],cnt,vis[Max];

void prim()

{

cnt=0;

int i,j;

memset(vis,0,sizeof(vis));

for(i=2;;i++)

{

if(!vis[i])

{

prime[++cnt]=i;

if(cnt>20010) return;

for(j=2*i;j<=Max;j+=i)

vis[j]=1;

}

}

}

void tostr(string &str,int x)

{

while(x)

{

str+=(x%10+'0');

x/=10;

}

}

int main()

{

// freopen("a.txt","r",stdin);

prim();

int n,i,j;

int a[15];

a[0] = 1;

for(i = 1; i <= 10; i ++)

a[i] = a[i-1] + (1<<(2*i-1));

while(scanf("%d",&n)==1)

{

string str="";

for(i=n;i>=1;i--)

tostr(str,prime[i]);

reverse(str.begin(),str.end());

int len = str.size();

for(i = 0; i <= 10; i ++)

if(len < a[i]) break;

cout << *(str.begin() +a[i-1] - 1) <<endl;

}

return 0;

}