输出区间 [l,r]上距离最近的相邻的素数对和距离最远的相邻的素数对。 请注意:3,5 是相邻的素数,2, 5不是相邻的素数。 ”距离”的定义为 2 个素数的差的绝对值
时间: 2024-10-08 22:21:18 浏览: 11
要解决这个问题,我们可以编写一个程序来找出给定区间[l, r]内的所有素数对,然后分别计算它们之间的最小距离(最近的)和最大距离(最远的)。首先,我们需要一个函数来判断一个数是否为素数:
```cpp
#include <iostream>
#include <vector>
#include <cmath>
bool isPrime(int num) {
if (num <= 1) return false;
if (num == 2 || num == 3) return true;
if (num % 2 == 0 || num % 3 == 0) return false;
for (int i = 5; i * i <= num; i += 6) {
if (num % i == 0 || num % (i + 2) == 0)
return false;
}
return true;
}
std::pair<int, int> findAdjacentPrimesPair(int l, int r) {
std::vector<int> primes;
for (int i = l; i <= r; ++i) {
if (isPrime(i))
primes.push_back(i);
}
if (primes.empty()) return {0, 0}; // No prime found in the range
int min_distance = INT_MAX;
int max_distance = 0;
for (size_t i = 1; i < primes.size(); ++i) {
int distance = std::abs(primes[i] - primes[i - 1]);
min_distance = std::min(min_distance, distance);
max_distance = std::max(max_distance, distance);
}
return {min_distance, max_distance};
}
int main() {
int l, r;
std::cout << "Enter the range (start and end): ";
std::cin >> l >> r;
auto result = findAdjacentPrimesPair(l, r);
std::cout << "The closest adjacent prime pair has a distance of: " << result.first << "\n";
std::cout << "The farthest adjacent prime pair has a distance of: " << result.second << "\n";
return 0;
}
```
这个程序首先遍历[l, r]找到所有的素数,然后计算连续素数之间的距离并记录最小和最大的距离。如果在指定范围内没有素数,返回距离为0和0。