传送门

题意： 给两个数$x,y$。现在你计算有多少对$a(a<=x)$和$b(b<=y)$使得$\lfloor \frac{a}{b}\rfloor =a\mathrm{m}\mathrm{o}\mathrm{d}b$。

思路： 因为$x$和$y$都是$1e9$的范围，可以想到$\sqrt{n}$求出答案。我们令$\lfloor \frac{a}{b}\rfloor =a\mathrm{m}\mathrm{o}\mathrm{d}b=k$，那么$a$可以写成$k\ast b+k$，又因为$k<b$，那么$k\ast k<=k\ast b+k=a<=x$，可得$k<=\sqrt{x}$，现在考虑知道了$k$能否算出答案呢？考虑如下三个个不等式：$$\{\begin{array}{l}1<=b<=y\\ 1<=k\ast b+k<=x\\ k<b\end{array}$$
解得：$$k<b<=min(y,x/k-1)$$
我们枚举$k$就可以得到答案啦。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int x,y;int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--){LL ans=0;scanf("%d%d",&x,&y);for(LL k=1;k*k<x;k++) ans+=max(0ll,min(1ll*x/k-1,1ll*y)-k);printf("%lld\n",ans);}return 0;
}
/**/