POJ 2689 Prime Distance 素数筛选法应用

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题目来源：POJ 2689 Prime Distance

题意：给出一个区间L R 区间内的距离最远和最近的2个素数并且是相邻的 R-L <= 1000000 但是L和R会很大

思路：一般素数筛选法是拿一个素数然后它的2倍3倍4倍...都不是然后这题可以直接从2的L/2倍开始它的L/2+1倍L/2+2倍...都不是素数

首先筛选出一些素数然后在以这些素数为基础在L-R上在筛一次因为 R-L <= 1000000 可以左移开一个1百万的数组

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1000010;typedef __int64 LL;int vis[maxn];int p[maxn];int prime[maxn];//筛素数 void sieve(int n){int m = sqrt(n+0.5);memset(vis, 0, sizeof(vis));vis[0] = vis[1] = 1;for(int i = 2; i <= m; i++)if(!vis[i])for(int j = i*i; j <= n; j += i)vis[j] = 1;}int get_primes(int n){sieve(n);int c = 0;for(int i = 2; i <= n; i++)if(!vis[i])prime[c++] = i;return c;}int main(){LL L, R;int c = get_primes(100000);while(scanf("%I64d %I64d", &L, &R) != EOF){if(L <= 2)L = 2;memset(p, 0, sizeof(p));for(int i = 0; i < c; i++){if(prime[i] > R)break;LL t = L / prime[i];if(t <= 1)t = 2;LL j = t*prime[i];while(j < L)j += prime[i];for(; j <= R; j += prime[i]){//printf("%d,,", j);p[j-L] = 1;}}LL ans1 = 999999999, ans2 = 0;LL x1, y1, x2, y2, x = -1, y = -1;for(LL i = L; i <= R; i++){if(!p[i-L]){//printf("%d\n", i);x = y;y = i;if(x != -1 && y != -1){if(ans1 > y-x){ans1 = y-x;x1 = x;y1 = y;}if(ans2 < y-x){ans2 = y-x;x2 = x;y2 = y;}}}}if(x == -1 || y == -1){puts("There are no adjacent primes.");continue;}printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n", x1, y1, x2, y2);}return 0;}

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