http://lightoj.com/volume_showproblem.php?problem=1341
The main topic: give you the area of the rectangle (the rectangle's side length are positive integers), let you find the smallest side is greater than the number of rectangles equal to B.
What is the unique decomposition theorem: The basic theorem of arithmetic can be expressed as: any one natural number greater than 1 N, if n is not prime , then n can be uniquely decomposed into a finite number of primes of the product N=p1a1p2a2p3A3 ... PnAn, where p1<p2<p3......<pn are prime, where the exponential AI is a positive integer. Such decomposition is called the standard decomposition of N
After we have found the number of factors of n, we divide by 2, get the number of factors in half, and then loop from the beginning to the B unqualified directly minus
#include <stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<iostream>#include<vector>#include<queue>using namespacestd;#defineN 1001000#defineESP 1e-8#defineINF 0x3f3f3f3f#defineMemset (A, B) memset (A,b,sizeof (a))intPrime[n], K, vis[n];voidPrime () {k=0; memset (Vis,0); for(intI=2; i<n; i++) { if(!Vis[i]) {Prime[k++] =i; for(intJ=i+i; j<n; j+=i) vis[j]=1; } }}///Prime number ScreeningLong LongSolveLong LongN) { Long Long intsum =1; for(intI=0; I<k && prime[i]*prime[i]<=n; i++) { if(N%prime[i] = =0) { intans=0; while(N%prime[i] = =0) {ans++; N/=Prime[i]; } Sum*= (1+ans); } } if(n>1) Sum*=2; returnsum;}///to find the number of n factors;intMain () {intT, t=1; scanf ("%d", &T); Prime (); while(T--) { Long Longb; scanf ("%lld%lld", &a, &b); if(A <= b*b) {printf ("Case %d:0\n", t++); Continue; } Long Long intnum =solve (a); Num/=2; for(intI=1; i<b; i++) { if(a% i = =0) Num--; } printf ("Case %d:%lld\n", t++, num); } return 0;}
1341-aladdin and the Flying Carpet---light oj (unique decomposition theorem + prime number screening)