Prime Distance

阅读原文时间：2023年07月08日阅读：34

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.
　　
题意：给定一个区间，找去这个区间里相邻的最近的和最远的两组素数。
思路：根据题目中的数据范围，可以看出要用到埃式筛法，直接暴力肯定是不行的，然后直接无脑写就行了，好吧这就是我调了两天才AC的原因，我用的是挑战程序设计竞赛的模板。。。。。。阿西吧。
一直过不了，不废话了，具体看代码中的注释吧。

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
10 #include
11 #include
12 #include
13 #include
14 #include
15 #include
16 #include
17 #include
18 #include
19 #define ci cin
20 #define co cout
21 #define el endl
22 #define Scc(c) scanf("%c",&c)
23 #define Scs(s) scanf("%s",s)
24 #define Sci(x) scanf("%d",&x)
25 #define Sci2(x, y) scanf("%d%d",&x,&y)
26 #define Sci3(x, y, z) scanf("%d%d%d",&x,&y,&z)
27 #define Scl(x) scanf("%I64d",&x)
28 #define Scl2(x, y) scanf("%I64d%I64d",&x,&y)
29 #define Scl3(x, y, z) scanf("%I64d%I64d%I64d",&x,&y,&z)
30 #define Pri(x) printf("%d\n",x)
31 #define Prl(x) printf("%I64d\n",x)
32 #define Prc(c) printf("%c\n",c)
33 #define Prs(s) printf("%s\n",s)
34 #define For(i,x,y) for(int i=x;iy;i--)
37 #define FFor_(i,x,y) for(int i=x;i>=y;i--)
38 #define Mem(f, x) memset(f,x,sizeof(f))
39 #define LL long long
40 #define ULL unsigned long long
41 #define MAXSIZE 1000005
42 #define INF 0x3f3f3f3f
43
44 const int mod=1e9+7;
45 const double PI = acos(-1.0);
46
47 using namespace std;
48
49 bool is_prime[MAXSIZE];
50 bool is_prime_small[MAXSIZE];
51 //is_prime[i-a]=true--->i是素数
52 void solve(LL a,LL b)
53 {
54 for(int i=0; (LL)i*i<=b; i++) 55 is_prime_small[i]=true; 56 //is_prime_small[0]=is_prime_small[1]=false; 57 for(int i=0; i<=b-a; i++) 58 is_prime[i]=true; 59 if(a==1) 60 is_prime[0]=0;//就是这个特判，我调了一天才搞出来。。。。。。 61 for(int i=2; (LL)i*i<=b; i++) 62 if(is_prime_small[i]) 63 { 64 for(int j=2*i; (LL)j*j<=b; j+=i) 65 is_prime_small[j]=false;//筛2~根号b 66 for(LL j=max(2LL,(a+i-1)/i)*i; j<=b; j+=i) 67 is_prime[j-a]=false;//筛a~b 68 } 69 } 70 //j = (ll)(a-1+i)/i*i 71 //(a-1+i)/i*i是对a/i向上取整,此计算的作用是求得第一个>=a的i的倍数。
72 //for(LL j=max(2LL,(a+i-1)/i)*i; j<=b; j+=i)这个循环的初始条件还是不懂 73 74 int main() 75 { 76 LL a,b; 77 LL c1,c2,d1,d2; 78 while(~Scl2(a,b)) 79 { 80 solve(a,b); 81 int tmp=-1; 82 int minn=INF,maxx=-1; 83 queueq1,q2;
84 For_(i,a,b)
85 {
86 if(is_prime[i-a])
87 {
88 if(tmp!=-1)
89 {
90 if(i-tmpmaxx)
97 {
98 maxx=i-tmp;
99 d1=tmp;
100 d2=i;
101 }
102 }
103 tmp=i;
104 }
105 }
106 if(minn!=INF||maxx!=-1)
107 cout<<c1<<","<<c2<<" are closest, "<<d1<<","<<d2<<" are most distant."<<endl;
108 else
109 Prs("There are no adjacent primes.");
110 }
111 return 0;
112 }

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