BZOJ1257 [CQOI2007] and the remainder of the [law]

Zcr and yy is a lot of ridicule in turn, felt IQ declining. . . T_T

Mod first split into integer division, and it is $ nk- \ Sigma_ {i = 1} ^ {n} i * \ lfloor \ frac {k} {i} \ rfloor $. Seek behind that.

And then found $ \ lfloor \ frac {k} {i} \ rfloor $ is continuous and monotonically growing. For $ x $, $ [x, \ lfloor \ frac {k} {\ lfloor \ frac {k} {i} \ rfloor} \ rfloor] $ within the same supplier. You can sense.

This is to find the law to get, and will not permit QWQ. So each segment as a quotient multiplied by each $ i $ this period of accumulation.

Complexity is $ i \ leq \ sqrt {k} have $ \ sqrt {k} $ $ when seed supplier. $ I \ geq \ sqrt {k} and Suppliers less than $ \ sqrt {k} $ $ time also up to $ \ sqrt {k} $ species. Thus the complexity is the root.

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WA: the right thing and every point of the interval to be sentenced on longlong exceeds $ n $.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 typedef long long ll;
 8 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 9 template<typename T>inline T read(T&x){
10     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
11     while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
12 }
13 inline ll sum(ll i,ll j){return (i+j)*(j-i+1)/2;}
14 ll ans,n,k;
15 
16 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
17     read(n),read(k);ans=n*k;(n>k)&&(n=k);
18     for(register ll i=1,r=0;i<=n;i=r+1)ans-=sum(i,r=_min(n,k/(k/i)))*(k/i);
19     return printf("%lld\n",ans),0;
20 }

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