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;}