C++(g++ 7.5.0) 解法, 执行用时: 89ms, 内存消耗: 408K, 提交时间: 2022-10-13 13:20:16

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll l, r, k;
int main() {
	cin>>l>>r>>k;
	ll res=0;
	for(ll L=1, R, dl, dr; L<=r; L=R+1) {
		dl=(l-1)/L;
		dr=r/L;
		R=r/dr;
		if(dl>0) R=min(R, (l-1)/dl);
		
		if(dr-dl>=k) res+=(R-L+1);
	}
	cout<<res;
	return 0;
}

C(gcc 7.5.0) 解法, 执行用时: 3ms, 内存消耗: 328K, 提交时间: 2022-10-01 09:39:59

#include<stdio.h>
int main()
{
	long long l,r,k,z,y;
	scanf("%lld%lld%lld",&l,&r,&k);
    l--;
	long long d=r-l,ans;
	ans=d/k;
	z=ans;
	long long t=l/ans;
	while(t>=0&&z<=d){
		y=r/(t+k);
		if(y>z)ans=ans+y-z;
		if(t!=0)z=l/t;
		t--;
	}
	printf("%lld\n",ans);
	return 0;
}

C++(clang++ 11.0.1) 解法, 执行用时: 101ms, 内存消耗: 496K, 提交时间: 2022-10-11 15:06:02

#include<bits/stdc++.h>
#define int long long
using namespace std;

signed main()
{
	int l,r,k;
	cin>>l>>r>>k;
	int s=0;
	l--;
	for(int x=1,y;x<=r;x=y+1)
	{
		if(x<=l) y=min(r/(r/x),(l/(l/x)));
		else y=r/(r/x);
		if(r/x-l/x>=k) s+=y-x+1;
	}
	cout<<s<<endl;
}