Aizu：0009- Prime Number

Prime Number

Time limit 1000 ms
Memory limit 131072 kB

Problem Description

Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.

Input

Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.

The number of datasets is less than or equal to 30.

Output

For each dataset, prints the number of prime numbers.

Sample Input

10
3
11

Output for the Sample Input

4
2
5

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解题心得：

1. 题意就是给你一个数n，问你在不大于n的数里有多少个素数。
2. 就是考了一个素数筛选，可以将素数找出来，然后二分查找n的位置，也可以直接得出前缀和。

---

二分查找

#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <vector>;
using namespace std;
const int maxn = 1e6+100;
bool prim[maxn];
vector <int> ve;

void get_prim() {
    prim[0] = prim[1] = true;
    for(int i=2;i<maxn;i++) {
        if(prim[i])
            continue;
        ve.push_back(i);
        for(int j=i*2;j<maxn;j+=i) {
            prim[j] = true;
        }
    }
}

int main() {
    int n;
    get_prim();
    while(scanf("%d",&n) != EOF) {
        int pos = upper_bound(ve.begin(),ve.end(),n) - ve.begin();
        if(ve[pos] > n)
            pos--;
        printf("%d
",pos+1);
    }
    return 0;
}

---

前缀和的代码

#include <algorithm>
#include <cstring>
#include <stdio.h>
using namespace std;
const int maxn = 1e6+100;
int sum[maxn];
bool prim[maxn];

void get_prim() {
    prim[1] = prim[0] = true;
    for(int i=2;i<maxn;i++) {
        if(prim[i])
            continue;
        for(int j=i*2;j<maxn;j+=i) {
            prim[j] = true;
        }
    }
}

void get_sum() {
    int cnt = 0;
    for(int i=0;i<maxn;i++) {
        if(!prim[i])
            cnt++;
        sum[i] = cnt;
    }
}

int main() {
    get_prim();
    get_sum();
    int n;
    while(scanf("%d",&n) != EOF) {
        printf("%d
",sum[n]);
    }
    return 0;
}