Prime Distance poj 2689 区间内的素数打表模板

Prime Distance Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10977   Accepted: 2957 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 先素数打表，把大范围的素数求出并标记，然后再把区间范围内的素数进行打表，再进行比较，求出最大和最小的范围素数 #include #include #include #include using namespace std; #define M 1000005 #define N 1 << 17 int l, u; bool is[N];//大范围的判断素数 long long prm[N];//大范围的素数打表 int n; bool ans[M];//判断区间内那些是素数 int prime[M];//把区间内素数记录下来 int num;//记录区间内的素数个数 int primed(int n) { int i, j, k = 0; int s, e = (int) (sqrt(0.0 + n) + 1); memset(is, 1, sizeof(is)); prm[k++] = 2; is[0] = is[1] = 0; for (i = 3; i < e; i += 2) if (is[i]) { prm[k++] = i; for (s = i * 2, j = i * i; j < n; j += s) is[j] = 0; } for (; i < n; i += 2) if (is[i]) prm[k++] = i; return k; } int main() { n = primed(1<<16);//记录共有多少个素数 while(scanf("%d%d",&l,&u)!=EOF) { for (int i = 0; i < u - l + 1; i++) ans[i] = true; if (l == 1) ans[0] = false; num = 0; /* for(int i=0; i<10; i++) { printf("prm %d\n",prm[i]); }*/ for (int i = 0; i < n && prm[i] * prm[i] <= u; i++) { long long temp = (l + prm[i] - 1) / prm[i] * prm[i]; /* if(i==0) { printf("a%lld b%lld c%lld %lld\n",temp,l + prm[i]- 1,prm[i],prm[i]*prm[i]); }*/ if (temp == prm[i]) temp += prm[i]; while (temp <= u) { ans[temp - l] = false; temp += prm[i]; } } /* for(int i=0; i prime[i + 1] - prime[i]) ans1 = i; if (prime[ans2 + 1] - prime[ans2] < prime[i + 1] - prime[i]) ans2 = i; } printf("%d,%d are closest, %d,%d are most distant.\n", prime[ans1], prime[ans1 + 1], prime[ans2], prime[ans2 + 1]); } } return 0; }