Идея: BledDest
Разбор
Tutorial is loading...
Решение (BledDest)
t = int(input())
for i in range(t):
n, m, k = map(int, input().split())
d = n // k
a1 = min(m, d)
a2 = (m - a1 + k - 2) // (k - 1)
print(a1 - a2)
Идея: BledDest
Разбор
Tutorial is loading...
Решение (pikmike)
t = int(input())
for _ in range(t):
n, m, x, y = map(int, input().split())
ans = 0
y = min(y, 2 * x)
for __ in range(n):
s = input()
i = 0
while i < m:
if s[i] == '*':
i += 1
continue
j = i
while j + 1 < m and s[j + 1] == '.':
j += 1
l = j - i + 1
ans += l % 2 * x + l // 2 * y
i = j + 1
print(ans)
Идея: adedalic
Разбор
Tutorial is loading...
Решение (pikmike)
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
int main() {
int tc;
scanf("%d", &tc);
forn(_, tc){
int h, c, t;
scanf("%d%d%d", &h, &c, &t);
if (h + c - 2 * t >= 0)
puts("2");
else{
int a = h - t;
int b = 2 * t - c - h;
int k = 2 * (a / b) + 1;
long long val1 = abs(k / 2 * 1ll * c + (k + 1) / 2 * 1ll * h - t * 1ll * k);
long long val2 = abs((k + 2) / 2 * 1ll * c + (k + 3) / 2 * 1ll * h - t * 1ll * (k + 2));
printf("%d\n", val1 * (k + 2) <= val2 * k ? k : k + 2);
}
}
return 0;
}
1359D - Yet Another Yet Another Task
Идея: BledDest
Разбор
Tutorial is loading...
Решение (pikmike)
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
const int INF = 1e9;
int main() {
int n;
scanf("%d", &n);
vector<int> a(n);
forn(i, n) scanf("%d", &a[i]);
long long ans = 0;
forn(mx, 31){
long long cur = 0;
long long best = 0;
forn(i, n){
int val = (a[i] > mx ? -INF : a[i]);
cur += val;
best = min(best, cur);
ans = max(ans, (cur - best) - mx);
}
}
printf("%lld\n", ans);
return 0;
}
Идея: BledDest
Разбор
Tutorial is loading...
Решение (BledDest)
#include<bits/stdc++.h>
using namespace std;
const int N = 500043;
const int MOD = 998244353;
int fact[N];
int add(int x, int y)
{
x += y;
while(x >= MOD) x -= MOD;
while(x < 0) x += MOD;
return x;
}
int mul(int x, int y)
{
return (x * 1ll * y) % MOD;
}
int binpow(int x, int y)
{
int z = 1;
while(y > 0)
{
if(y % 2 == 1)
z = mul(z, x);
x = mul(x, x);
y /= 2;
}
return z;
}
int inv(int x)
{
return binpow(x, MOD - 2);
}
int divide(int x, int y)
{
return mul(x, inv(y));
}
void precalc()
{
fact[0] = 1;
for(int i = 1; i < N; i++)
fact[i] = mul(i, fact[i - 1]);
}
int C(int n, int k)
{
if(k > n) return 0;
return divide(fact[n], mul(fact[n - k], fact[k]));
}
int main()
{
int n, k;
cin >> n >> k;
int ans = 0;
precalc();
for(int i = 1; i <= n; i++)
{
int d = n / i;
ans = add(ans, C(d - 1, k - 1));
}
cout << ans << endl;
}
Идея: BledDest
Разбор
Tutorial is loading...
Решение (pikmike)
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
#define x first
#define y second
using namespace std;
const double INF = 1e13;
struct line{
int A, B, C;
line(){}
line(int x1, int y1, int x2, int y2){
A = y1 - y2;
B = x2 - x1;
C = -A * x1 - B * y1;
// A is guaranteed to be non-zero
if (A < 0) A = -A, B = -B, C = -C;
int g = __gcd(A, __gcd(abs(B), abs(C)));
A /= g, B /= g, C /= g;
}
};
bool operator ==(const line &a, const line &b){
return a.A == b.A && a.B == b.B && a.C == b.C;
}
double x;
struct comp{
bool operator ()(const pair<line, int> &a, const pair<line, int> &b) const {
double val1 = (-a.x.A * x - a.x.C) / a.x.B;
double val2 = (-b.x.A * x - b.x.C) / b.x.B;
if (val1 != val2)
return val1 < val2;
return a.y < b.y;
}
};
struct car{
int x, y, dx, dy, s;
line l;
double vx, vy;
};
int n;
vector<car> a(n);
long long det(int a, int b, int c, int d){
return a * 1ll * d - b * 1ll * c;
}
bool inter(const line &a, const line &b, long long &D, long long &Dx, long long &Dy){
D = det(a.A, a.B, b.A, b.B);
if (D == 0) return false;
Dx = -det(a.C, a.B, b.C, b.B);
Dy = -det(a.A, a.C, b.A, b.C);
return true;
}
int sg(int x){
return x < 0 ? -1 : 1;
}
int sg(long long a, long long b, int c){
// sign of a/b-c
if (b < 0) a = -a, b = -b;
return a - c * b < 0 ? -1 : (a - c * b > 0);
}
bool inter(int i, int j, double &len){
if (i == -1 || j == -1)
return false;
long long D, Dx, Dy;
if (!inter(a[i].l, a[j].l, D, Dx, Dy))
return false;
if (sg(Dx, D, a[i].x) != 0 && sg(a[i].dx) != sg(Dx, D, a[i].x))
return false;
if (sg(Dx, D, a[j].x) != 0 && sg(a[j].dx) != sg(Dx, D, a[j].x))
return false;
double x = Dx / double(D);
double y = Dy / double(D);
double di = sqrt((a[i].x - x) * (a[i].x - x) + (a[i].y - y) * (a[i].y - y));
double dj = sqrt((a[j].x - x) * (a[j].x - x) + (a[j].y - y) * (a[j].y - y));
return len >= max(di / a[i].s, dj / a[j].s);
}
vector<char> del;
set<pair<line, int>, comp> q;
void get_neighbours(int i, int &l, int &r){
l = r = -1;
auto it = q.lower_bound({a[i].l, -1});
while (it != q.end() && del[it->y])
it = q.erase(it);
if (it != q.end())
r = it->y;
while (!q.empty() && it != q.begin()){
--it;
if (del[it->y])
it = q.erase(it);
else{
l = it->y;
break;
}
}
}
bool check(double t){
vector<pair<double, pair<int, int>>> cur;
del.assign(n, 0);
forn(i, n){
double x1 = a[i].x;
double x2 = a[i].x + a[i].vx * t;
if (x1 > x2) swap(x1, x2);
cur.push_back({x1, {i, 0}});
cur.push_back({x2, {i, 1}});
}
q.clear();
sort(cur.begin(), cur.end());
for (auto &qr : cur){
x = qr.x;
int i = qr.y.x;
int l, r;
if (qr.y.y == 0){
get_neighbours(i, l, r);
if (r != -1 && a[i].l == a[r].l)
return true;
if (inter(i, l, t))
return true;
if (inter(i, r, t))
return true;
q.insert({a[i].l, i});
}
else{
del[i] = true;
get_neighbours(i, l, r);
if (inter(l, r, t))
return true;
}
}
return false;
}
int main() {
scanf("%d", &n);
a.resize(n);
forn(i, n){
scanf("%d%d%d%d%d", &a[i].x, &a[i].y, &a[i].dx, &a[i].dy, &a[i].s);
a[i].l = line(a[i].x, a[i].y, a[i].x + a[i].dx, a[i].y + a[i].dy);
double d = sqrt(a[i].dx * a[i].dx + a[i].dy * a[i].dy);
a[i].vx = a[i].dx / d * a[i].s;
a[i].vy = a[i].dy / d * a[i].s;
}
double l = 0, r = INF;
bool ok = false;
forn(_, 100){
double m = (l + r) / 2;
if (check(m)){
ok = true;
r = m;
}
else{
l = m;
}
}
if (!ok)
puts("No show :(");
else
printf("%.15lf\n", l);
return 0;
}