UVALive6050 Primes【素数筛选+前缀和】

Primes

Let m and n be two integers, 2 ≤ m < n ≤ 10000000. Consider the following set:
P rime(m, n) = {p|p prime and m ≤ p ≤ n}.
Compute the cardinal of the set P rime(m, n).
Input
The input consists of several tests. The input of each test is represented on a single line in the input.
Any two consecutive tests are separated by an empty line. For each test, the values for m and n are
given on the same line, separated by exactly one space.
Output
For each test, the result will be written to standard output on a different line (the tests will have the
same order as in the input file). The results of any two consecutive tests will be separated by an empty
line. For each test, the result will be the cardinal of the set Prime(m, n).
Sample Input
2 20
70 110
5 150
Sample Output
8
10
33

Regionals 2012 >> Europe - Southeastern

问题链接：UVALive6050 Primes。

题意简述：参见上文。

问题分析：用Eratosthenes筛选法筛选素数，然后计算素数数量的前缀和，根据输入的数据范围做个减法即可。

程序说明：程序中有一个坑，需要注意。

参考链接：（略）

AC的C++语言程序如下：

/* UVALive6050 Primes */

#include <iostream>
#include <string.h>
#include <math.h>

using namespace std;

const int N = 10000000;
int prefixsum[N+1];

// Eratosthenes筛选法+计算前缀和
void sieveofe(int n)
{
    memset(prefixsum, 0, sizeof(prefixsum));

    prefixsum[0] = prefixsum[1] = 1;
    for(int i=2; i<=sqrt(n); i++) {
        if(!prefixsum[i]) {
            for(int j=i*i; j<=n; j+=i)  //筛选
                prefixsum[j] = 1;
        }
    }

    prefixsum[0] = 0;
    for(int i=1; i<=n; i++)
        if(prefixsum[i])
            prefixsum[i] = prefixsum[i - 1];
        else
            prefixsum[i] = prefixsum[i - 1] + 1;
}

int main()
{
    sieveofe(N);

    int n, m;
    bool flag = false;

    while(cin >> n >> m) {
        if(n>m)             // 坑：没说哪个数大
            swap(n, m);
        if(flag)
            cout << endl;
        cout << prefixsum[m] - prefixsum[n - 1] << endl;
        flag = true;
    }

    return 0;
}