[POJ 2689] Prime Distance【欧拉筛】

  • 2017-12-30
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

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.

Source

Waterloo local 1998.10.17

Solution:

用线性筛筛出 [2, sqrt(maxInt)] 上的素数,并同时维护出 [L, U] 上的素数(在数组 isrprm[] 中时将坐标偏移 -L 保存)。

注意以下几点:

  1. long long 存储防止累加溢出
  2. L == 1 时,防止将 1 当作素数处理。

Code: O(TR), 其中R=10^6 [1236K, 125MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxInt = 1000000007;

ll prm[25005];
int prmSize;
bool isprm[50005];

ll L, U;
bool isrprm[1000005];

inline void calcPrime(){
	memset(isprm, 1, sizeof(isprm));
	memset(isrprm, 1, sizeof(isrprm));
	isprm[0] = isprm[1] = 0, prmSize = 0;
	if(L == 1) isrprm[0] = 0;
	// Beware of the ignorance of regarding 1 as a prime!!!
	for(register int i = 2; i <= 50000; i++){
		if(isprm[i]) prm[++prmSize] = i;
		for(register int j = 1; j <= prmSize && i * prm[j] <= 50000; j++){
			isprm[i * prm[j]] = 0;
			if(i % prm[j] == 0) break;  // The core of Euler's Sieve
		}  // Primes below sqrt(maxInt) are enough
		for(register ll j = i * max(2LL, (L - 1) / i + 1); j <= U; j += i) isrprm[j - L] = 0;
		// Update primes in [L, U] synchronously, using "long long" to avoid overflowing
	}
}
int main(){
	while(scanf("%lld%lld", &L, &U) != EOF){
		calcPrime();
		pair<ll, ll> cls(-maxInt, maxInt), dst(-maxInt, -maxInt);  // Record the answers
		ll prev = -1;
		for(register ll i = L; i <= U; i++){
			if(!isrprm[i - L]) continue;
			if(prev != -1){
				if(i - prev < cls.second - cls.first) cls.second = i, cls.first = prev;
				if(i - prev > dst.second - dst.first) dst.second = i, dst.first = prev;
			}
			prev = i;
		}
		if(cls.first == -maxInt) puts("There are no adjacent primes.");
		else printf("%lld,%lld are closest, %lld,%lld are most distant.\n", cls.first, cls.second, dst.first, dst.second);
	}
	return 0;
}

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OPEN AT 2017.12.10

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