POJ 2689 区间素数筛选

题意：给区间[L,U]，要求该区间内，距离最小的素数对，和距离最大的素数对，输出素数对和对应距离。（距离的定义：2个相邻素数的差的绝对值）

分析：

1、筛选法

2、由于区间L,U (1<=L< U<=2,147,483,647)本身可能很大，直接筛选数组都开不下。

　但是题目说“The difference between L and U will not exceed 1,000,000.”也就是说U-L<=1000 000,这就给我们启示只筛选区间内部的素数。

　　然后由于区间是一个连续区间，所以我们用x(L=<x<=U)%1000000就可以把x映射到1000 000内的数了。这样开一个1000 000的数组存素数就好（当然，区间内的素数肯定没这么多）

3、注意用long long因为筛选过程可能超int（嘛……如果用int的话，本地都应该跑不出来才对吧……前提是自己随便输一个大一点的数……我也是程序死循环了才改的……T T）

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<map>
 5 #include<string>
 6 using namespace std;
 7 const int MAXN = 50001;
 8 const int MOD = 1000001;
 9 int prime[MAXN];
10 int isprime[MAXN*20];
11 int prim[MAXN*20];
12 int init(){
13   int k = 0;
14   memset(prime,0,sizeof prime);
15   for(int i = 2;i<MAXN;i++){
16     //cout<<"
"<<i<<endl;
17     if(!prime[i]){
18       prime[k++] = i;
19       long long j = (long long )i*(long long )i;
20       //cout<<j<<endl;
21       for(;j<MAXN;j+=i)prime[j] = 1;
22     }
23   }
24   return k;
25 }
26 int shaixuan(long long L,long long U,int pn){
27   memset(isprime,0,sizeof isprime);
28   //cout<<prime[0]<<"
U:"<<U<<endl;
29   for(long long i = 0;prime[i]<U+1 && i<pn;i++){
30     //cout<<"???
";
31     long long tp = L/prime[i];
32     if(tp < 2) tp = (long long )prime[i]*(long long )prime[i];
33     else{
34       tp = tp*(long long )prime[i];
35       if(tp < L)tp+=(long long )prime[i];
36     }
37     //cout<<prime[i]<<" "<<tp<<"
";
38     //if(!isprime[tp%MOD]){
39       //cout<<tp%MOD<<endl;
40       for(long long j = tp;j<U+1;j+=(long long )prime[i]){
41         isprime[j%MOD] = 1;
42       }
43     //}
44   }
45   int k = 0;
46   for(long long i = L;i<U+1;i++){
47     if(i == 1)continue;
48     if(!isprime[i%MOD]){
49       prim[k++] = i;
50       //cout<<i<<endl;
51     }
52   }
53   return k;
54 }
55 int main()
56 {
57 #ifdef LOCALL
58 //    freopen("in.txt","r",stdin);
59   //  freopen("out.txt","w",stdout);
60 #endif
61   int pn = init();
62   int L,U;
63   while(cin>>L>>U){
64     if(U == 1 || U == 2){cout<<"There are no adjacent primes."<<endl;continue;}
65     int pn2 = shaixuan(L,U,pn);
66     if(pn2 == 1){cout<<"There are no adjacent primes."<<endl;continue;}
67     long long minn ,maxn = 0,minx,maxx;
68     minx = maxx = 0;
69     minn = 1<<30;
70     //cout<<pn2<<endl;
71     //cout<<minn<<endl;
72     for(int i = 1;i<pn2;i++){
73      // cout<<"prim["<<i<<"]:"<<prim[i]<<endl;
74       //cout<<"sub:"<<prim[i]-prim[i-1]<<endl;
75       if(prim[i]-prim[i-1] < minn) {minn = prim[i]-prim[i-1]; minx = i-1;}
76       if(prim[i]-prim[i-1] > maxn) {maxn = prim[i]-prim[i-1]; maxx = i-1;}
77     }
78 
79     //cout<<"min:"<<minn<<" maxx:"<<maxn<<"maxx:"<<maxx<<endl;
80     cout<<prim[minx]<<","<<prim[minx+1]<<" are closest, "<<prim[maxx]<<","<<prim[maxx+1]<<" are most distant.
";
81 
82   }
83   return 0;
84 }

POJ2689

总之，筛选法简直打素数表之万能神器……