UVa 10539 (Sieve primes, two-point search) almost prime Numbers

Test instructions

Find the number of integers between L and U x satisfies the form such as X=PK, where P is prime, k>1

Analysis:

First, sift out the primes in 1e6, enumerate each prime number to find out 1e12 of all the numbers that satisfy the condition, and then sort.

For L and U, the two points find the subscript of the maximum number less than U and L, and the difference is the answer.

1#include <cstdio>2#include <cmath>3#include <algorithm>4 5typedefLong LongLL;6 7 Const intMAXN =1000000;8 Const intMAXP =80000;9 Ten BOOLVIS[MAXN +Ten]; One intPRIME[MAXP], CNTP =0; ALL A[MAXN], cnt =1; -  - voidInit () the { -     intm = +; -      for(inti =2; I <= m; ++i)if(!Vis[i]) -          for(intj = i * I; J <= Maxn; J + = i) vis[j] =true; +      for(inti =2; I <= MAXN; ++i)if(!vis[i]) prime[cntp++] =i; -  +      for(inti =0; i < CNTP; ++i) A     { atll temp = (LL) prime[i] *Prime[i]; -          while(Temp <=1000000000000LL) -         { -a[cnt++] =temp; -Temp *=Prime[i]; -         } in     } -Std::sort (A, A +CNT); to } +  - intBinary_search (LL N) the { *LL L =0, R =CNT; $      while(L <R)Panax Notoginseng     { -LL mid = ((L + R +1) >>1); the         if(A[mid] <= N) L =mid; +         ElseR = mid-1; A     } the     returnL; + } -  $ intMain () $ { - Init (); -     intT; thescanf"%d", &T); -      while(t--)Wuyi     { the LL hehe, haha; -scanf"%lld%lld", &hehe, &haha); Wu         intt = Binary_search (hehe), k =binary_search (haha); -         intAns = k-T; About         if(A[t] = = hehe) ans++; $printf"%d\n", ans); -     } -  -     return 0; A}

code June

UVa 10539 (Sieve primes, two-point search) almost prime Numbers