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）
1 var n,k,i,l,r,m:longint; 2 ans:qword; 3 function ceil(x:real):longint; 4 begin 5 if trunc(x)<x then exit(trunc(x)+1) else exit(trunc(x)); 6 end; 7 procedure main; 8 begin 9 readln(n,k); 10 ans:=0; 11 if n>k then 12 begin 13 inc(ans,(n-k)*k); 14 n:=k; 15 end; 16 m:=ceil(sqrt(k)); 17 for i:=1 to m do inc(ans,k mod i); 18 for i:=1 to m do 19 begin 20 l:=(k div (i+1))+1; 21 r:=(k div i); 22 if l<=m then l:=m+1; 23 if r>n then r:=n; 24 if r<l then continue; 25 inc(ans,(((k<<1)-i*(l+r))*(r-l+1)>>1)); 26 end; 27 writeln(ans); 28 end; 29 begin 30 main; 31 end.
代码：（c++）
1 #include <iostream> 2 #include <cmath> 3 using namespace std; 4 5 long long ans,n,k; 6 int main() { 7 ios::sync_with_stdio(false); 8 cin>>n>>k; 9 if (n>k) { 10 ans+=k*(n-k); 11 n=k; 12 } 13 long long sqrtk=ceil(sqrt(k)); 14 for (long long i=1;i<=sqrtk;++i) ans+=k%i; 15 for (long long a=1;a<=sqrtk;++a) { 16 long long L=floor(k/(a+1))+1; 17 long long R=floor(k/a); 18 if (L<=sqrtk) L=sqrtk+1; 19 if (R>n) R=n; 20 if (R<L) continue; 21 ans+=((k<<1)-a*L-a*R)*(R-L+1)>>1; 22 } 23 cout<<ans; 24 }
另一种分块方法，pascal还是过不了……
1 #include <cmath> 2 #include <cstdio> 3 #include <algorithm> 4 5 long long n, k, ans; 6 int main() 7 { 8 scanf("%lld%lld", &n, &k); 9 if (n > k) 10 { 11 ans += (n-k)*k; 12 n = k; 13 } 14 ans += n * k; 15 for (long long i = 1, last; i <= n; i = last+1) 16 { 17 last = std::min(n, k/(k/i)); 18 ans -= (k/i) * (i+last) * (last-i+1) / 2; 19 } 20 printf("%lld", ans); 21 }