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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#87440#5503. Euclidean AlgorithmchirankoTL 18514ms7284kbC++141.1kb2023-03-12 21:45:422023-03-12 21:45:45

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-03-12 21:45:45]
  • 评测
  • 测评结果:TL
  • 用时:18514ms
  • 内存:7284kb
  • [2023-03-12 21:45:42]
  • 提交

answer

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
#define rep(l,r) for(int i=l;i<=r;i++)
#define jrep(l,r) for(int j=l;j<=r;j++)

using namespace std;

typedef __int128 lll;
typedef long long ll;
const int P=500000,N=P+10;

ll ff[N];
bitset<N> st;

ll f(ll n)
{
	if(n<=P&&st[n])return ff[n];
	ll cnt=n>=2;
	ll l=1,r;
	n--;
	while(l<=n)
	{
		ll t=n/(double)l;
		r=n/(double)t;
		cnt+=(r-l+1ll)*(t-1);
		l=r+1;
	}
	if(n<P)
	{
		st[n+1]=true;
		return ff[n+1]=cnt;
	}
	return cnt;
}

int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);
	int T;cin>>T;
	while(T--)
	{
		ll n;cin>>n;
		ll ans=0;
		ll l=1,r;
		while(l<=n>>1)
		{
			ll t=n/(double)l;
			r=min(n>>1,ll(n/(double)t));
			ans+=(r-l+1ll)*(t-2);
			l=r+1;
		}
		// cout<<ans<<' ';
		l=1;
		while(l<=n)
		{
			ll t=n/(double)l;
			r=n/(double)t;
			ans+=(r-l+1ll)*f(t);
			l=r+1;
		}
		cout<<ans<<'\n';
	}
	return 0;
}

/*
3
100000000000
100000000000
100000000000

*/
/*
3
1000000000
1000000000
1000000000
*/

详细

Test #1:

score: 100
Accepted
time: 2ms
memory: 3468kb

input:

3
2
5
14

output:

1
9
62

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 18514ms
memory: 7284kb

input:

3
29107867360
65171672278
41641960535

output:

8921593237533
21300450379732
13136180138425

result:

ok 3 lines

Test #3:

score: -100
Time Limit Exceeded

input:

3
90076809172
100000000000
99913139559

output:

30192292781431
33790187414013

result: