POJ2689 Prime Distance

时间限制: 1 Sec  内存限制: 512 MB
提交: 104  解决: 16
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题目描述

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers. 
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

输入

Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.

 

输出

For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.

 

样例输入

复制样例数据

2 17
14 17

样例输出

2,3 are closest, 7,11 are most distant.
There are no adjacent primes.

显然求出[1,r]中所有的素数是不现实的，但r-l比较小，可以求出[l,r]间的素数。
素数不好求，可以先求出合数，不是合数的就是素数。任何一个合数n必然包括一个不超过sqrt(n)的质因子，先打一个素数表，然后判断合数。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e6 + 100;
const int N = sqrt(2147483647.0);
ll v[maxn], prime[maxn];
int cnt = 0;
int e[maxn];
int vis[maxn];

void get_prime() {
    for (int i = 2; i <= N; i++) {
        if (!v[i]) {
            v[i] = i;
            prime[++cnt] = i;
        }
        for (int j = 1; j <= cnt; j++) {
            if (prime[j] > v[i] || prime[j] > N / i) break;
            v[i * prime[j]] = prime[j];
        }
    }
}

int main() {
    //freopen("input.txt", "r", stdin);
    int l, r;
    get_prime();
    while (scanf("%d %d", &l, &r) != EOF) {
        int tot = 0;
        int pos = 1;
        memset(vis, 0, sizeof(vis));
        if (1 - l >= 0) vis[1 - l] = 1;//要把1加进去
        while (prime[pos] <= r && pos <= cnt) {
            int from = l / prime[pos];
            int to = r / prime[pos];
            from = max(from, 2);
            for (int i = from; i <= to; i++) {
                if (prime[pos] * i - l >= 0 && prime[pos] * i - l < maxn)
                    vis[prime[pos] * i - l]++;
            }
            pos++;
        }
        for (int i = 0; i <= r - l; i++) {
            if (!vis[i]) {
                e[tot++] = i + l;
            }
        }
        if (tot < 2) {
            printf("There are no adjacent primes.
");
            continue;
        }
        int minl, minr, mi = 0x3f3f3f3f, maxl, maxr, ma = -1;
        for (int i = 1; i < tot; i++) {
            if (e[i] - e[i - 1] < mi) {
                mi = e[i] - e[i - 1];
                minl = e[i - 1];
                minr = e[i];
            }
            if (e[i] - e[i - 1] > ma) {
                ma = e[i] - e[i - 1];
                maxl = e[i - 1];
                maxr = e[i];
            }
        }
        printf("%d,%d are closest, %d,%d are most distant.
", minl, minr, maxl, maxr);

    }
    return 0;
}