POJ 2689 Sieve method to calculate prime number

DES: Give a range [L, U]. Find the nearest two primes in this interval and the two prime numbers farthest away from each other. The length of 1<=l<u<=2147483647 interval is not more than 1000000.

Idea: Because the given u range exceeds the maximum of int. So you can't directly hit the 1-u prime table. "We know. When using the prime sieve method, all the composite are sifted away. Then we can also find a way to l-u inside the composite are filtered out. So as long as the sqrt (U) in the prime number to sift out the l-u composite can be. " just about to understand ....

1#include <stdio.h>2#include <iostream>3#include <string.h>4#include <iostream>5 using namespacestd;6 7 #defineN 5000008 #defineLen 100000009 #defineINF 0x7fffffffTen BOOLisprime[n+5]; One Long LongPrime[n], CNT; A BOOLres[len+5]; -  - voidInit () {//Filter the prime number within 50000. Be aware that I will be more prone to the range of int when it is larger the      Long LongI, J; -CNT =0; -memset (IsPrime,true,sizeof(IsPrime)); -       for(i=2; i<=n; ++i) { +         if(Isprime[i]) { -prime[cnt++] =i; +             if(I*i <=N) { A                  for(J=i*i; j<=n; j+=i) { atISPRIME[J] =false; -                 } -             } -         } -      } - } in  - intMain () { to     Long LongL, U, I, J, K, Minn, Maxx, S, T; + init (); -      while(~SCANF ("%lld%lld", &l, &U)) { thememset (Res,0,sizeof(res)); *          for(i=0; i<cnt; ++i) { $s = (l1)/prime[i]+1;//s and T represent the minimum and maximum multiples of the number pairs for this prime in the current range. For example, the interval [3, 9] for the prime number 2 S and T should be 2, 4Panax Notoginsengt = U/prime[i];//By This example can know why L-1 finally again +1. T can be removed directly.  -              for(j=s; j<=t; ++j) {//filter out the primes in the interval the                 if(j>1) {//Do not know why ==1 time also not calculate?????  +RES[J*PRIME[I]-L] =true; A                 } the             } +         } -K =-1, Minn = inf, Maxx =-1; $         Long LongDIS, m1, M2; $          for(i=0; i<=u-l; ++i) {//Solving optimal values -             if(!Res[i]) { -                 if(k!=-1) { theDis=i-k;//record the maximum distance of two primes.  -                     if(Dis >Maxx) {WuyiMaxx =dis; theM1 =i; -                     } Wu                     if(Dis <Minn) { -Minn =dis; AboutM2 =i; $                     } -                 } -                 if(I+l! =1)//Note that when you are 1, the special Judgment 1 is not prime.  -k=i; A             } +         } the         if(Maxx = =-1) -printf"there is no adjacent primes.\n"); $         Elseprintf"%lld,%lld is closest,%lld,%lld is most distant.\n", M2-minn+l, M2+l, M1-maxx+l, m1+L); the     } the     return 0; the}

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The code can be understood. Naked knocks like ... It's still a bit difficult. I'll knock it again tomorrow.

POJ 2689 Sieve method to calculate prime number