首页 > 代码库 > POJ-2689 Prime Distance(线性筛法)
POJ-2689 Prime Distance(线性筛法)
Prime Distance
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17021 | Accepted: 4536 |
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).
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
个人认为,线性筛法的精髓之处在于:比一个合数数大的质数和该合数的乘积可用一个更大的合数和比其小的质数相乘得到。
嗯
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define N 50000 5 #define len 1000000 6 #define inf 0x7fffffff 7 using namespace std; 8 typedef long long LL; 9 bool isprime[N+5]; 10 LL prime[N],cnt; 11 bool res[len+5]; 12 void init(){ 13 LL i,j; 14 cnt=0; 15 memset(isprime,true,sizeof(isprime)); 16 for (i=2;i<=N;i++){ 17 if (isprime[i]) prime[cnt++]=i; 18 for (j=0;j<cnt && i*prime[j]<=N;j++){ 19 isprime[i*prime[j]]=false; 20 if (i%prime[j]==0) break; 21 } 22 } 23 } 24 int main(){ 25 freopen ("prime.in","r",stdin); 26 freopen ("prime.out","w",stdout); 27 LL L,U,i,j,k,min,max,s,t; 28 init(); 29 while(scanf("%lld%lld",&L,&U)!=EOF){ 30 memset(res,0,sizeof(res)); 31 for(i=0;i<cnt;i++){ 32 s=(L-1)/prime[i]+1; 33 t=U/prime[i]; 34 for(j=s;j<=t;j++) 35 if(j>1) 36 res[j*prime[i]-L]=true; 37 } 38 k=-1,min=inf,max=-1; 39 LL dis,m1,m2; 40 for(i=0;i<=U-L;i++){ 41 if(!res[i]){ 42 if(k!=-1){ 43 dis=i-k; 44 if(dis>max){ 45 max=dis; 46 m1=i; 47 } 48 if(dis<min){ 49 min=dis; 50 m2=i; 51 } 52 } 53 if(i+L!=1)//注意 1 的时候特殊判断 1 不是素数 54 k=i; 55 } 56 } 57 if(max==-1) 58 printf("There are no adjacent primes.\n"); 59 else{ 60 printf("%lld,%lld are closest, %lld,%lld are most distant.\n",m2- 61 min+L,m2+L,m1-max+L,m1+L); 62 } 63 } 64 return 0; 65 }
POJ-2689 Prime Distance(线性筛法)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。