1561. PRIME

1561. PRIME

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. The first prime number is 2. Can you write a program that computes the nth prime number, given a number n <= 10000?

Input

The input contains just one number which is the number n as described above.
The maximum value of n is 10000.

Output

The output consists of a single line with an integer that is the nth prime number.

Sample Input

30

Sample Output

113

【解题思路】

本题主要是求第N个素数，经典之处在于如何减少资源利用率，计算到第10000，坏的算法是不可能再一秒钟之内结束的，肯定会超时。

判定n为素数，即他不被 根号n 内的素数整除即为素数。

然后将已判定的素数存在vector中，以防止重复判定。

最后以0.05sec，312kb的运行内存计算完成，这个数据相对理想、

【遇到的问题】

判定素数的时候break过多，用错位置导致

另计数从1开始，而非从0开始，但是却有vector[0]。

【源代码】

#include<iostream>
#include<vector> 
#include <math.h>
#include <stdlib.h>
using namespace std;

int main()
{
    vector <int> prime;
    int i=5;
    prime.push_back(2);
    prime.push_back(3);
    prime.push_back(5);
    for(;;)
    {
           i++;
           for(int j=0;;j++)
           {
                  if(i % prime[j] == 0)break;                  
                  if(prime[j] > sqrt(i)+1)
                  {
                       prime.push_back(i);
                       break;
                  }
           }
            
    if(prime.size()>10000)break;
    }  
    int index = 0;
    cin>>index;
    cout<<prime[index-1]<<endl;
    return 0;
}