UVa 10791 (unique decomposition) Minimum Sum LCM

Test instructions

Enter N to find at least two positive integers, so that the least common multiple of these numbers is n and minimum.

Analysis:

The decomposition of the n is, apparently alone, as an item, and the smallest.

Here are two tips:

• Starting from 2, the constant addition of n, until it cannot be divisible. This eliminates the problem of prime judgment and shortens the amount of code. Because the first 2 of all n is out of the factor, then there will be no n 4 = = 0 of the case, so that all except n is a prime number.
• From 2 in addition to n until sqrt (n), if n is not 1, then the "remaining" is at this time n the largest mass factor. Reduce the number of cycles.

1#include <cstdio>2#include <cmath>3 4typedefLong LongLL;5 6 intMainvoid)7 {8     intN, Kase =0;9      while(SCANF ("%d", &n) = =1&&N)Ten     { OneLL ans =0; A         intm = sqrt (n +0.5); -         intpcnt =0; -          the         if(n = =1) -         { -printf"Case %d:2\n", ++Kase); -             Continue; +         } -          +          for(inti =2; I <= m; ++i) A         { at             if(n% i = =0) -             { -pcnt++; -                 inttemp =1; -                  while(n% i = =0) -                 { inN/=i; -Temp *=i; to                 } +                 if(Temp >1) ans + =temp; -             } the         } *         if(N >1) $         {Panax Notoginsengpcnt++; -Ans + =N; the         } +         if(Pcnt <=1) ans++; A          theprintf"Case %d:%lld\n", ++Kase, ans); +          -     } $      $     return 0; -}

code June

UVa 10791 (unique decomposition) Minimum Sum LCM