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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
#139620#5503. Euclidean Algorithmammardab3anML 1ms3476kbC++203.0kb2023-08-14 05:02:092023-08-14 05:02:10

Judging History

你现在查看的是最新测评结果

  • [2023-08-14 05:02:10]
  • 评测
  • 测评结果:ML
  • 用时:1ms
  • 内存:3476kb
  • [2023-08-14 05:02:09]
  • 提交

answer


// By AmmarDab3an 

#include "bits/stdc++.h"

using namespace std;

#define int int64_t
#define ll  int64_t

// typedef unsigned int        uint;
// typedef long long int       ll;
// typedef unsigned long long  ull;
typedef pair<int, int>    pii;
typedef pair<ll, ll>      pll;
typedef pair<int, pii>    iii;
typedef pair<ll, pll>     lll;
typedef vector<int>       vi;
typedef vector<ll>        vl;
typedef vector<pii>       vpii;
typedef vector<pll>       vpll;

#define endl '\n'
#define fastIO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define freopenI freopen("input.txt", "r", stdin);
#define freopenO freopen("output.txt", "w", stdout);

const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const double  PI = acos(-1);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
int rand(int x, int y) {
	return uniform_int_distribution<int>(x, y)(rng);
}

int mul(int a, int b){
	int ret = (1ll * (a%MOD) * (b%MOD)) % MOD;
	return (ret+MOD)%MOD;
}
 
int add(int a, int b){
	int ret = (1ll * (a%MOD) + (b%MOD)) % MOD;
	return (ret+MOD)%MOD;
}
 
int pow_exp(int n, int p){
	if(!p) return 1;
	if(p&1) return mul(n, pow_exp(n, p-1));
	int tmp = pow_exp(n, p/2);
	return mul(tmp, tmp);
}

int inv(int x){
	return pow_exp(x, MOD-2);
}
 
const int  MAX = 2e5 + 10;
const int NMAX = 2e5 + 10;
const int MMAX = 2e5 + 10;
const int LOG_MAX = ceil(log2(double(NMAX)));
const int BLOCK = ceil(sqrt(double(NMAX)));

int fac[NMAX], ifac[NMAX];

void init(){
	
	fac[0] = 1;
	for(int i = 1; i < NMAX; i++){
		fac[i] = mul(fac[i-1], i);
	}
	
	ifac[NMAX-1] = inv(fac[NMAX-1]);
	for(int i = NMAX-2; i >= 0; i--){
		ifac[i] = mul(ifac[i+1], i+1);
	}
}

int choose(int n, int c){
	assert(n >= c);
	return mul(fac[n], mul(ifac[c], ifac[n-c]));
}

long long solve(long long n) 
{
    long long s = sqrtl(n);
    long long ret = 0;
    for (int i = 1; i <= s; ++ i)
    {
        ret += n / i;
    }
    ret *= 2; 
    ret -= s * s; 
    return ret;
}

int brute_force(int n){
	
	int ans = 0;
	
	for(int i = 1; i <= n; i++)
	for(int g = 1; g <= n; g++){
		
		if(g*i > (n-g)) break;
		
		// g*(k*i+1) <= n
		// k*i+1 <= n/g
		// k*i <= n/g -1
		// k <= (n/g -1) / i
		// k <= (n-g)/(g*i)
		
		ans += (n-g) / (g*i);
		
		// for(int k = 1; g*(k*i+1) <= n; k++){
			// ans++;
		// }
	}
		
	return ans;
}

int32_t main(){
    
    fastIO;
    
#ifdef LOCAL
    freopenI;
    freopenO;
#endif

    // freopen("name.in", "r", stdin);
    
	// init();
	
    int t; cin >> t; while(t--){

		int n;
		cin >> n;
		
		vpii tmp;
		
		int cur = n;
		int pst = 0;
		
		while(cur >= 1){
		  int cnt = n/cur - pst;
		  tmp.push_back({cur, cnt});
		  pst = n/cur;
		  cur = n/(n/cur + 1);
		}
		
		int ans = 0;
		
		for(auto [a, b] : tmp){
			ans += b * solve(a-1);
		}
		
		// cout << brute_force(n) << endl;
		cout << ans << endl;
    }	
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3476kb

input:

3
2
5
14

output:

1
9
62

result:

ok 3 lines

Test #2:

score: -100
Memory Limit Exceeded

input:

3
29107867360
65171672278
41641960535

output:


result: