UVA10852 Less Prime【筛选法】

Let n be an integer, 100 ≤ n ≤ 10000, find the prime number x, x ≤ n, so that n − p ∗ x is maximum, where p is an integer such that p ∗ x ≤ n < (p + 1) ∗ x.
Input
The first line of the input contains an integer, M, indicating the number of test cases. For each test case, there is a line with a number N, 100 ≤ N ≤ 10000.
Output
For each test case, the output should consist of one line showing the prime number that verifies the condition above.
Sample Input
5
4399
614
8201
101
7048
Sample Output
2203
311
4111
53
3527

问题链接：UVA10852 Less Prime
问题简述：给定一个整数n，求一个素数x，满足p*x ≤ n < (p+1)x，且n - px 最大。
问题分析：
    简单题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10852 Less Prime */

#include <bits/stdc++.h>

using namespace std;

const int N = 10000;
bool isprime[N + 1];
int prime[N / 3], pcnt = 0, maxm[N  + 1];
// 欧拉筛
void eulersieve(void)
{
    memset(isprime, true, sizeof(isprime));

    isprime[0] = isprime[1] = false;
    for(int i = 2; i <= N; i++) {
        if(isprime[i])
            prime[pcnt++] = i;
        for(int j = 0; j < pcnt && i * prime[j] <= N; j++) {  //筛选
            isprime[i * prime[j]] = false;
            if(i % prime[j] == 0) break;
        }
    }

    for (int i = 2 ; i <= N ; i++) {
        maxm[i] = prime[0];
        for (int j = 1 ; j < pcnt ; j++) {
            if (prime[j] >= i) break;
            if (i % maxm[i] < i % prime[j])
                maxm[i] = prime[j];
        }
    }
}

int main()
{
    eulersieve();

    int m, n;
    scanf("%d", &m);
    while(m--) {
        scanf("%d", &n);
        printf("%d\n", maxm[n]);
    }

    return 0;
}

来源：CSDN

作者：海岛Blog

链接：https://blog.csdn.net/tigerisland45/article/details/104551057