LA 3521 Joseph ' s problem

Test instructions: give you positive integers n and k, and then calculate the and from I to n k%i;

idea; if n is less than 1000000, the direct violence is calculated, then the case is greater than 1000000, and then in the discussion of the size of N and K, according to the k%i situation, you will find that the rule is multiple arithmetic progression, and then you add these arithmetic progression to the answer.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 #definell Long Long5 using namespacestd;6 7 ll N,k;8 ll Getsum (ll N)9 {Tenll sum=0; One      for(LL i=1; i<=n; i++) A     { -sum+=k%i; -     } the     returnsum; - } -  - intMain () + { -      while(SCANF ("%lld%lld", &n,&k)! =EOF) +     { A         if(n<=1000000) at         { -printf"%lld\n", Getsum (n)); -             Continue; -         } -ll ans=0; -Ans+=max (LL)0, n-k) *K; in          for(intI=2; i<=10000; i++) -         { to             if(i>k) Break; +ll x1=k/(i-1)-k/i; -             if(k/i>n)Continue; the             ints=k% (k/(i-1)), e=k% (k/i+1); *             if(k/(i1) >N) ${s=k%N;Panax Notoginsengx1=n-k/i; -             } theans+= (s+e) *x1/2; +         } A         if(k>10000) the         { +ll m=k/10000; -ans+=getsum (m); $         } $printf"%lld\n", ans); -     } -     return 0; the}

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LA 3521 Joseph ' s problem