led question nds state rip input bsp round easy

Description

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).

Input

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.

Output

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.

Sample Input

2 17
14 17

Sample Output

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

解題思路：

這題做得我都是淚。不斷地TLE，好不easy優化好了。又RE。代碼也寫得非常齪。就是正常的二次篩選素數。

因為數據非常大。第一次篩出46500以內的素數。再依據此篩選出區間內的素數。

註意：盡管給的數沒有超int範圍，但兩數相乘是會超int範圍的，我也是在這裏RE了。

AC代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1000005;
const int M = 46500;
const int INF = 999999999;
bool notprime[N];
int prime_1[M + 1], prime_2[N];
int num_1 = 0, num_2;
void Prime1()  // 第一次篩出46500以內的素數
{
    memset(notprime, false, sizeof(notprime));
    for(int i = 2; i <= M; i++)
        if(!notprime[i])
        {
            prime_1[num_1++] = i;
            for(int j = 2 * i; j <= M; j += i)
                notprime[j] = true;
        }
}
void Prime2(int l, int u)  // 第二次篩出給定範圍內的素數
{
    memset(notprime, false, sizeof(notprime));
    num_2 = 0;
    if(l < 2)
        l = 2;
    int k = sqrt(u * 1.0);
    for(int i = 0; i < num_1 && prime_1[i] <= k; i++)
    {
        int t = l / prime_1[i];
        if(t * prime_1[i] < l)
            t++;
        if(t <= 1)
            t = 2;
        for(int j = t; (long long)j * prime_1[i] <= u; j++)  // 相乘會超範圍，用long long
            notprime[j * prime_1[i] - l] = 1;
    }
    for(int i = 0; i <= u - l; i++)
        if(!notprime[i])
            prime_2[num_2++] = i + l;
}
int main()
{
    int l, u, dis, a_1, b_1, a_2, b_2, minn, maxx;;
    Prime1();
    while(scanf("%d%d", &l, &u) != EOF)
    {
        minn = INF, maxx = -1;
        Prime2(l, u);
        if(num_2  < 2)
        {
            printf("There are no adjacent primes.\n");
            continue;
        }
        for(int i = 1; i < num_2 && prime_2[i] <= u; i++)
        {
            dis = prime_2[i] - prime_2[i - 1];
            if(dis > maxx)
            {
                a_1 = prime_2[i - 1];
                a_2 = prime_2[i];
                maxx = dis;
            }
            if(dis < minn)
            {
                b_1 = prime_2[i-1];
                b_2 = prime_2[i];
                minn = dis;
            }
        }
        printf("%d,%d are closest, %d,%d are most distant.\n", b_1, b_2, a_1, a_2);
    }
    return 0;
}

Prime Distance（二次篩素數）