Prime Distance

  1 #include <map>
  2 #include <set>
  3 #include <list>
  4 #include <stack>
  5 #include <queue>
  6 #include <deque>
  7 #include <cmath>
  8 #include <ctime>
  9 #include <string>
 10 #include <limits>
 11 #include <cstdio>
 12 #include <vector>
 13 #include <iomanip>
 14 #include <cstdlib>
 15 #include <cstring>
 16 #include <istream>
 17 #include <iostream>
 18 #include <algorithm>
 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;i<y;i++)
 35 #define For_(i,x,y) for(int i=x;i<=y;i++)
 36 #define FFor(i,x,y) for(int i=x;i>y;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         queue<int>q1,q2;
 84         For_(i,a,b)
 85         {
 86             if(is_prime[i-a])
 87             {
 88                 if(tmp!=-1)
 89                 {
 90                     if(i-tmp<minn)
 91                     {
 92                         minn=i-tmp;
 93                         c1=tmp;
 94                         c2=i;
 95                     }
 96                     if(i-tmp>maxx)
 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 }