CQOIX2007余数之和

朴素能得个差不多吧……

这题改进算法真恶心

pascal一直过不了，难道非得转c++？

代码：（pascal）

 var n,k,i,l,r,m:longint;
     ans:qword;
 function ceil(x:real):longint;
  begin
  if trunc(x)<x then exit(trunc(x)+) else exit(trunc(x));
  end;
 procedure main;
  begin
    readln(n,k);
    ans:=;
    if n>k then
     begin
      inc(ans,(n-k)*k);
      n:=k;
     end;
    m:=ceil(sqrt(k));
    for i:= to m do inc(ans,k mod i);
    for i:= to m do
     begin
       l:=(k div (i+))+;
       r:=(k div i);
       if l<=m then l:=m+;
       if r>n then r:=n;
       if r<l then continue;
       inc(ans,(((k<<)-i*(l+r))*(r-l+)>>));
     end;
    writeln(ans);
  end;
 begin
   main;
 end.    

代码：（c++）

 #include <iostream>
 #include <cmath>
 using namespace std;

 long long ans,n,k;
 int main() {
   ios::sync_with_stdio(false);
   cin>>n>>k;
   if (n>k) {
     ans+=k*(n-k);
     n=k;
   }
   long long sqrtk=ceil(sqrt(k));
   for (long long i=;i<=sqrtk;++i) ans+=k%i;
   for (long long a=;a<=sqrtk;++a) {
     long long L=floor(k/(a+))+;
     long long R=floor(k/a);
     if (L<=sqrtk) L=sqrtk+;
     if (R>n) R=n;
     if (R<L) continue;
     ans+=((k<<)-a*L-a*R)*(R-L+)>>;
   }
   cout<<ans;
 }

另一种分块方法，pascal还是过不了……

 #include <cmath>
 #include <cstdio>
 #include <algorithm>

 long long n, k, ans;
 int main()
 {
   scanf("%lld%lld", &n, &k);
   if (n > k)
   {
     ans += (n-k)*k;
     n = k;
   }
   ans += n * k;
   for (long long i = , last; i <= n; i = last+)
   {
     last = std::min(n, k/(k/i));
     ans -= (k/i) * (i+last) * (last-i+1) / 2;
   }
   printf("%lld", ans);
 }