标签：des style blog io ar color os 使用 sp
The sequence of n ? 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, 24, 25, 26, 27, 28 between 23 and 29 is a prime gap of length 6.
Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.
The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.
The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.
Copy sample input to clipboard 10 11 27 2 492170 0
4 0 6 0 114
分析：首先不可能输入每个数都去找一遍其周围的素数，我比较喜欢的是直接求出所需要的最大范围的素数，然后从这里面再进行下一步的计算。这里使用的是筛选法来得到素数集。
#include <iostream> using namespace std; int main(int argc, char const *argv[]) { int prime[1299710]; for (int i = 2; i != 1299710; ++i) { prime[i] = 1; } for (int i = 2; i != 1299710; ++i) { if (prime[i] == 1) { for (int j = 2 * i; j < 1299710; j += i) prime[j] = 0; } } // 筛选法找素数 int num; while (cin >> num && num != 0) { int length = 0; if (prime[num] != 1) { int upperBound = 0, lowerBound = 0; for (lowerBound = num - 1; lowerBound >= 2; --lowerBound) { if (prime[lowerBound] == 1) break; } for (upperBound = num + 1; upperBound < 1299710; ++upperBound) { if (prime[upperBound] == 1) break; } length = upperBound - lowerBound; } cout << length << endl; } return 0; }
标签：des style blog io ar color os 使用 sp
原文地址：http://www.cnblogs.com/xiezhw3/p/4132892.html
