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name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int str[2000000];
bool can[2000000];
int main() {
int i, j, k;
int n, m, d;
int now;
while (cin >> n >> d) {
for (i = 1; i <= n; i++) {
scanf("%d", &str[i]);
}
sort(str + 1, str + 1 + n);
now = 0;
memset(can, false, sizeof(can));
can[0]... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
int dp[51 * 10001];
int main(void) {
int n, d, a, sum, i, j;
while (scanf("%d%d", &n, &d) != EOF) {
sum = 0;
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for (i = 0; i < n; i++) {
scanf("%d", &a);
sum += a;
for (j = sum; j >= a; j--) {
if (dp[j - a] == 1) ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int a[500003];
vector<int> p;
int main() {
int N, D, u;
cin >> N >> D;
for (int i = 0; i < 500003; ++i) {
a[i] = 0;
}
a[0] = 1;
p.push_back(0);
for (int i = 0; i < N; ++i) {
cin >> u;
for (int j = p.size(); j--;) {
if (p[j] + u < 500003 && !a... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
bitset<500005> dp;
int N, D, ans, i, j, x, d, s = 1;
int main() {
dp[0] = 1;
for (scanf("%d%d", &N, &D), i = 0; i < N; i++) {
scanf("%d", &x);
for (j = 500000; j >= x; j--) dp[j] = (dp[j] || dp[j - x]);
}
while (s) {
for (s = 0, i = min(ans + D, 500000);... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int N = 55;
const int D = 10100;
int dp[N * D];
int n, d;
int A[N];
void do_package() {
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for (int i = 0; i < n; i++) {
for (int j = N * D - 1; j >= 0; j--) {
if (dp[j] && j + A[i] < N * D) {
dp[j + A[i]] = 1... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int dp[550001], c[10005];
int main() {
int n, d;
cin >> n >> d;
for (register int i = 1; i <= n; i++) {
cin >> c[i];
}
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for (register int i = 1; i <= n; i++) {
for (register int j = 550000; j >= c[i]; j--) {
i... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int dp[5000100];
int main() {
int i, j, n, d;
int cnt = 0, sum = 0, c = 0;
dp[0] = 1;
cin >> n >> d;
int p;
for (i = 0; i < n; i++) {
cin >> p;
for (j = (sum += p); j >= p; j--)
if (dp[j - p] == 1) dp[j] = 1;
}
while (1) {
j = d + c;
wh... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const unsigned long long P = 239017, MaxN = 2100000, INF = 1000000000;
int n, d, can[510100], a[1000];
vector<int> q;
int main() {
scanf("%d%d", &n, &d);
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
can[0] = 1;
for (int i = 0; i < n; ++i) {
for (int have = 50... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 500010;
bool vis[MAXN] = {0};
int M[MAXN];
int N, K, T;
int main() {
cin >> N >> K;
vis[0] = true;
for (int i = 1; i <= N; i++) {
cin >> T;
for (int j = i * 10000; j >= 0; j--)
if (vis[j]) vis[j + T] = true;
}
int all = 0;
for (int... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int sz = 112345;
int a[100];
bool dp[52][500005];
int sums[500005];
int main() {
int n, d, i, j, ans, day;
scanf("%d", &n);
scanf("%d", &d);
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
for (i = 0; i < 500005; i++) dp[0][i] = 0;
dp[0][0] = 1;
fo... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int dp[555555];
int a[55];
int n, k;
int main() {
memset(dp, 0, sizeof(dp));
cin >> n >> k;
for (int i = 0; i < n; i++) cin >> a[i];
dp[0] = 1;
for (int i = 0; i < n; i++)
for (int j = 555555; j >= a[i]; j--) dp[j] |= dp[j - a[i]];
int ans = 0, day = 0;
wh... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout << fixed;
cout.precision(12);
solve();
return 0;
}
template <typename T>
void sc(T& x) {
cin >> x;
}
template <typename Head, typename... Tail>
void sc(Head& head, Tail&... tail) {
cin ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
vector<int> arr;
int n, d;
int have;
int lastHave;
int dp[50][10000 * 55];
bool knapsack() {
for (int j = lastHave + 1; j <= have + d; ++j)
dp[0][j] = (arr[0] <= j) ? arr[0] : 0;
for (int i = 1; i < n; ++i) {
for (int j = lastHave + 1; j <= have + d; ++j) {
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int arr[50 + 5];
bool Sum[(int)1e4 * 50 + 5];
int main() {
int N, D, ans = 0, maxi = 0, tot = 0;
scanf("%d %d", &N, &D);
for (int i = 0; i < N; i++) {
scanf("%d", &arr[i]);
tot += arr[i];
}
Sum[0] = 1;
for (int i = 0; i < N; i++)
for (int j = tot - a... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int N, D;
int val[60];
int dp[10010 * 60];
int main() {
int i, j;
cin >> N >> D;
int sum = 0;
dp[0] = 1;
for (i = 0; i < N; i++) {
cin >> val[i];
sum += val[i];
for (j = sum; j >= val[i]; j--) {
if (dp[j - val[i]]) {
dp[j] = 1;
}
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6;
int a[maxn], dp[maxn];
int main() {
int n, d;
scanf("%d%d", &n, &d);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
dp[0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 600000; j >= a[i]; j--) dp[j] = max(dp[j], dp[j - a[i]]);
int ans... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:268435456,268435456")
using namespace std;
int a[50];
bool dp[510001];
int main() {
int n, d;
scanf("%d %d", &n, &d), dp[0] = 1;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
for (int j = 510001 - 1; j >= a[i]; j--) dp[j] |= dp[j - a[i]];
}
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int N = 555555;
bool dp[N];
int main() {
int n, d;
while (cin >> n >> d) {
memset(dp, 0, sizeof(dp));
int c;
dp[0] = 1;
for (int i = 0; i < n; i++) {
cin >> c;
for (int j = N - 1; j >= c; j--) dp[j] |= dp[j - c];
}
int mx = 0, c... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const long long A = 100000000000000LL, N = 2228228;
long long dp[N], i, j, n, r, m, a, o[2];
int main() {
cin >> n >> m, dp[0] = 1;
for (i = 0; i < n; i++) {
cin >> a, r += a;
for (j = r; j >= a; j--)
if (dp[j - a]) dp[j] = 1;
}
while (1) {
j = o[0... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | import java.util.*;
public class b {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int n = input.nextInt(), d = input.nextInt();
boolean[] possible = new boolean[1000000];
possible[0] = true;
int[] data = new int[n];
for(int i = 0; i<n; i++) data[i] = input.nex... | JAVA |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int a[300], b[500001], m, n, d;
bool used[51][500001], used1[500001];
void solve(int i, int s) {
used1[s] = true;
if (i > n) return;
if (used[i][s]) return;
used[i][s] = true;
solve(i + 1, s);
solve(i + 1, s + a[i]);
}
int main() {
cin >> n >> d;
for (int i ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
int _MINUS;
const int MXN = 2000010;
const int C = 262144;
const int INF = 1000000001;
const int MXS = 500010;
bool knapsack[MXN];
int n, d;
int c[MXN];
priority_queue<pair<int, int> ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 511111;
int num[maxn];
int q[maxn];
int main() {
int i, j, n, m, k;
scanf("%d%d", &n, &m);
memset(num, 0, sizeof(num));
num[0] = 1;
for (i = 1; i <= n; i++) {
scanf("%d", &k);
for (j = 500000; j >= k; j--) {
if (num[j - k]) num[j] = ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int MN = 50;
const int MVAL = 500000;
const int MM = 505050;
bool can[MM];
int main() {
int n, d;
scanf("%d %d", &n, &d);
can[0] = true;
int i, j;
for ((i) = 0; (i) < (int)(n); ++(i)) {
int a;
scanf("%d", &a);
for (j = MVAL; j >= 0; --j) {
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
int n, m, d;
bool dp[55 * 10100];
int main() {
dp[0] = true;
scanf("%d %d", &n, &d);
for (int i = 0; i < n; ++i) {
int t;
scanf("%d", &t);
for (int j = m; j >= 0; --j)
if (dp[j]) dp[j + t] = true;
m += t;
}
int cur = 0, days = 0;
bool changed = true;
while (c... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 55, maxv = 500005;
int n, D, a[maxn], f[maxv], b[maxv], now, cnt;
int main() {
scanf("%d%d", &n, &D);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
f[0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 500000; j >= a[i]; j--) f[j] |= f[j - a[i]... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int maxix = 60;
const int maxnum = 1000 * 1000 * 1000 + 10;
const int maxd = 50 * 10000 + 10000;
int dp[maxd], mark[maxd];
int arr[maxix];
int main() {
int n, d;
cin >> n >> d;
for (int i = 1; i <= n; i++) {
cin >> arr[i];
}
sort(arr + 1, arr + n + 1);
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int M = 500000 + 10;
int a[M];
int dp[2][M];
vector<int> x;
int main() {
int n, d;
cin >> n >> d;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
dp[0][0] = 1;
for (int i = 1; i <= n; i++) {
int flag = i % 2;
int pre = 1 - flag;
for (int j = ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int p[50 * 10000 + 1];
int v[50];
int solve_problem() {
vector<int> positions;
int n, d;
if (scanf("%d %d", &n, &d) != 2) return 1;
for (int i = 0; i < n; i++)
if (scanf("%d", &v[i]) != 1) return 1;
int s = 0;
for (int i = 0; i < n; i++) s += v[i];
sort(v,... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout << fixed;
cout.precision(12);
solve();
return 0;
}
template <typename T>
void sc(T& x) {
cin >> x;
}
template <typename Head, typename... Tail>
void sc(Head& head, Tail&... tail) {
cin ... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int dp[500001];
int main() {
int n, d, x = 0, i, j;
int a[50];
scanf("%d %d", &n, &d);
for (i = 0; i < n; i++) scanf("%d", &a[i]);
dp[0] = 1;
for (i = 0; i < n; i++) {
for (j = 500000; j >= 0; j--) {
if (dp[j] == 1) dp[j + a[i]] = 1;
}
}
for (i... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int n, d, a[55], f[500005][55], s[1000005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cin >> n >> d;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
fill(f[0], f[0] + n + 2, 1);
vector<int> v(1);
for (int i = 1; i < 500005; i++) {
for (... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
stringstream ss;
long long mod = 1000000007LL;
int a[64], dp[500010];
int main() {
int n, d;
cin >> n >> d;
dp[0] = 1;
for (int i = 0; i < n; i++) {
cin >> a[i];
for (int j = 500000; j >= a[i]; j--)
if (dp[j - a[i]]) dp[j] = 1;
}
vector<int> cand;
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const double EPS = 0.0000001;
const double PI = acos(-1);
const long long INFLL = 0x7FFFFFFFFFFFFFFF;
const int INF = 0x7FFFFFFF;
template <typename T>
inline void next(T &num) {
char c;
num = 0;
do {
c = getchar_unlocked();
} while (c != EOF && c == ' ' && c ==... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int main() {
set<int> s;
s.insert(0);
int n, d;
cin >> n >> d;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
for (int i = 0; i < n; i++) {
set<int> temp;
for (set<int>::iterator it = s.begin(); it != s... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int q[505050], f[505050], g[505050], a[505050];
int main() {
int n, D;
scanf("%d%d", &n, &D);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
int m = 0;
for (int i = 1; i <= n; i++) m += a[i];
memset(g, 0, sizeof(g));
g[0] = 1;
for (int i = 1; i <= n; i++)
... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
bool used[50];
int n;
int ss[500001], v[50];
int main(void) {
int d;
int i, ans = 0, sum = 0, j;
scanf("%d%d", &n, &d);
for (i = 0; i < n; i++) {
scanf("%d", &v[i]);
sum += v[i];
}
ss[0] = true;
for (i = 0; i < n; i++)
for (j = sum; j >= v[i]; j--)... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:268435456,268435456")
using namespace std;
bool dp[510001];
int main() {
int n, d, t;
scanf("%d %d", &n, &d), dp[0] = 1;
for (int i = 0; i < n; i++) {
scanf("%d", &t);
for (int j = 510000; j - t >= 0; j--) dp[j] |= dp[j - t];
}
int curr = 0, num... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 10000 * 55;
const long long LINF = 1000000000LL * 1000000000LL;
bool f[MAXN], fn[MAXN];
long long ff[MAXN];
int i, j, n, w, d, a[111];
map<long long, int> x;
int main() {
cin >> n >> d;
for (i = 0; i < n; i++) {
cin >> a[i];
w += a[i];
}
f[0... | CPP |
365_D. Free Market | John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.
John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that eac... | 2 | 10 | #include <bits/stdc++.h>
using namespace std;
int n, d, a[60], t;
bool best[500100];
vector<int> s;
int main() {
cin >> n >> d;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
t += a[i];
}
best[0] = 1;
for (int i = 1; i <= n; ++i)
for (int j = t - a[i]; j >= 0; --j) {
best[j + a[i]] = max(best[j... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long INF = 1e18;
int main() {
double n, i, k, m = 101;
cin >> n;
vector<long long> a(n, 0);
vector<double> b(101, 0);
for (i = 0; i < (n); i++) {
cin >> a[i];
b[a[i]]++;
}
double x = 0, y = 0;
for (i = 0; ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long i, j, k, l, n, a, b, ans = 0;
int main() {
std::ios_base::sync_with_stdio(0);
cin >> n;
vector<long long> num(n);
for (i = 0; i < n; i++) {
cin >> num[i];
}
sort(num.begin(), num.end());
for (i = 0; i < n; i++) {
if (num[i] == -1) continue;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long a;
cin >> a;
vector<long long> x;
for (long long i = 0; i < a; i++) {
long long b;
cin >> b;
x.push_back(b);
}
bool done[101] = {false};
sort(x.begin(), x... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
vector<int> v(n);
for (long long int i = 0; i < n; i++) cin >> v[i];
sort(v.begin(), v.end());
int ans = 0;
while (!v.empty()) {
ans++;
int num = 0;
for (lon... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import sys
inp = [map(int, i.split(' ')) for i in sys.stdin.read().strip().splitlines()]
dat = sorted(inp[1])
res = 0
while dat:
chain = 0
for i in xrange(len(dat)):
if dat[i] >= chain:
chain += 1
dat[i] = -1
res += 1
dat = filter(lambda x: x >= 0, dat)
sys.stdout.wri... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUF... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class CSolver {
private static InputReader in;
private static PrintWriter out;
public static void main(String[] args) thro... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;
import java.io.InputStream;
import java.io.InputStreamReader;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static ... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool v[110];
int cnt[110];
vector<int> a;
int main() {
int test;
int n;
cin >> n;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
if (v[x] == false) {
v[x] = true;
}
a.push_back(x);
cnt[x]++;
}
int l = a.size();
vector<int> v1;
s... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | input();print(max([((ind+1)//(i+1)+((ind+1)%(i+1)!=0)) for ind,i in enumerate(sorted(map(int,input().split())))])) | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;import java.io.*;import java.math.*;
public class Main
{
public static void process()throws IOException
{
int n=ni();
int[]A=nai(n);
Arrays.sort(A);
int ans=1;
for(int i=1;i<n;i++)
if(A[i]<(i/ans))
ans++;
pn(ans);
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.FileReader;
import java.io.IOException;
import java.math.BigInteger;
import java.text.ParseException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
public class CodeForces {
class pairT implements Comparable<pairT>{
int count;
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool used[110];
int main() {
int n, a[110], ans = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
while (true) {
int i;
for (i = 0; i < n; i++)
if (!used[i]) break;
if (i == n) break;
ans++;
int h = 0;
for (i = 0... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
static void solve() throws Exception {
int n = nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
Arrays.sort... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[105], b[105];
bool cmp(int x, int y) { return x > y; }
int main() {
int t, i, n, j;
cin >> n;
for (i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + 1, a + 1 + n, cmp);
for (t = 1;; t++) {
memset(b, -1, sizeof(b));
bool flag = 0;
for (i ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n;
int a[101];
int b[101];
vector<int> x;
int res;
void Init() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
b[a[i]]++;
}
}
void AB() {
for (int i = 0; i <= 100; i++) {
while (b[i] > 0 && i >= x.size()) {
x.push_back(i);
b[i]--;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = raw_input()
n = int(n)
boxes = raw_input()
boxes = boxes.split()
boxes = map(int, boxes)
piles = []
while len(boxes) > 0:
pile = []
s = max(boxes)
boxes.remove(s)
if s != 0:
temp_boxes = boxes
while len(temp_boxes) > 0 and len(pile)<s:
smallest = min(temp_boxes)
... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | '''
Created on 01-Jul-2017
@author: kandarp
'''
import sys
def check(x,a):
if x ==0 :
return False;
b =[]
ans = True;
for i in xrange(0,x):
b.append([]);
for i in xrange(0,len(a)):
b[i%x].append(a[i])
for i in xrange(0,x):
for j in xrange(0,len(b[i])):
... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.Arrays;
import java.util.Scanner;
public class FoxAndBox {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] x = new int[n];
for(int i=0; i<n; i++){
x[i] = sc.nextInt();
}
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
li=list(map(int,input().split(" ",n)[:n]))
li.sort()
ans=1
for i in range(n):
if li[i]< i//ans:
ans+=1
print(ans)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
Outpu... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class cf388A {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int boxes = sc.nextInt();
int[] strengths = new int[101];
for(int i = 0; i < boxes; i++) {
strengths[sc.nextInt()]++;
}
int[] piles =... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, count = 0;
int arr[200];
cin >> n;
for (int i = 0; i < n; i++) cin >> arr[i];
sort(arr, arr + n);
vector<int> ans;
int k = 0;
int l;
ans.push_back(1);
for (int i = 1; i < n; i++) {
if (arr[i] == 0)
ans.push_back(1);
else... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import math
from decimal import *
getcontext().prec=30
n=int(input())
x=list(map(int,input().split()))
x.sort()
l=list()
for i in range(len(x)):
added=0
for j in range(len(l)):
if x[i]>=len(l[j]):
l[j].append(x[i])
added=1
break
if not added:
v=[x[i]]
... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
int b[110], vis[110];
int cmp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
int main(void) {
int n, i, cur, k, pile;
while (scanf("%d%*c", &n) != EOF) {
for (i = 0; i < n; i++) scanf("%d", &b[i]);
qsort(b, n, 4, cmp);
memset(vis, 0, sizeof(vis));
k = 0;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.io.*;
//DONT FORGET TO CHANGE CLASS NAME!
public class FoxAndBoxAccumulation {
/*
First, we need to put all of our box strengths into an array.
We can then sort the array from least to greatest
Finally, we have a linked list to keep track of the number of boxes in each p... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public final class FoxBox {
/**
* @param args
*/
public static void main(String[] args) {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
try ... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
vector<pair<long long, pair<long long, long long> > > v;
for (long long i = 1; i <= n; i++) {
long long x;
cin >> x;
v.push_back(make_pair(x, make_pair(1, x)));
}
sort(v.begin(), v.end());
while (1) {
bool pa... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, t, ans;
multiset<pair<int, int> > S;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> t;
S.insert(make_pair(t, 1));
}
while (!S.empty()) {
bool F = false;
pair<int, int> cur = *S.begin();
S.erase(S.begin());
for (multiset<pa... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | from collections import defaultdict
def partition(seq, key=int):
d = defaultdict(int)
for x in seq:
d[key(x)] += 1
return d
def popBox(boxes, x):
if boxes[x] == 1:
del boxes[x]
else:
boxes[x] -= 1
nBoxes = int(raw_input())
boxes = partition([x for x in raw_input().split()]... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Stack;
import java.util.StringTokenizer;
import static java.util.Arrays.sort;
public class FoxAndBox... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(raw_input())
X = map(int, raw_input().split())
L = [[] for i in xrange(100)]
X.sort()
for i in xrange(n):
x = X[i]
for j in xrange(n):
if len(L[j]) <= x:
L[j] += [x]
break
print sum([L[i] != [] for i in xrange(n)]) | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | def Acc(arr):
arr.sort()
arr_aux=[0 for i in range(len(arr))]
i=0
j=0
while i<len(arr):
if arr[i]>=arr_aux[j]:
arr_aux[j]=arr_aux[j]+1
i+=1
j=0
else:
j+=1
#print(arr_aux)
valor_actual=len(arr)-arr_aux.count(0)
print(valor_ac... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(input())
l = list(map(int,input().split(' ')))
l.sort()
nb = 0
for i in range(n):
if l[i] == -1 :
continue
h = 1
nb += 1
for j in range(i+1,n):
if l[j]>=h:
h+=1
l[j] = -1
print(nb) | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | """Template for Python Competitive Programmers prepared by Mayank Chaudhary aka chaudhary_19"""
# to use the print and division function of Python3
from __future__ import division, print_function
"""value of mod"""
MOD = 10 ** 9 + 7
"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
int main() {
auto start = high_resolution_clock::now();
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<int> v(n);
for (int i = 0; i < n; ++i) cin >> v[i];
sort(v.begin(), v.end());
priority_queue<in... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
int main() {
int sayac = 0, x, n, t1, t = 0, i, j, arr[105] = {0};
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &x);
arr[x]++;
}
while (1) {
t = 0;
t1 = 1;
for (i = 0; i <= 100; i++)
if (arr[i]) t1 = 0;
if (t1) break;
for (i = 0; i <= 100; i... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=input()
h=[0]
p=sorted(map(int,raw_input().split()))
for x in p:
if x<min(h):
h+=[1]
else:
h[h.index(min(h))]+=1
print len(h) | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
std::cin >> n;
int arr[n];
for (int i = 0; i < n; i++) {
std::cin >> arr[i];
}
sort(arr, arr + n);
vector<int> vis(n, 0);
int ans = 0, total = n;
while (total) {
int cnt = 0;
for (int i = 0; i < n; i++) {
if (!vis[i] &... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #from collections import deque
n = map(int,input().split())
x = sorted(list(map(int,input().split())))
piles = 1
current = x.pop(0)
current_size = 1
while x:
for i,e in enumerate(x):
if e >= current_size:
current = e
x.pop(i)
current_size += 1
break
else:... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class FoxAndBox {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = Integer.parseInt(s.nextLine());
List<Integer> boxes = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
boxes.add(s.nextInt());
}
Collections.sort(boxes);
int count = 0;
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int arr[1000], ans = 0;
int cnt[1000];
int main() {
int n;
scanf("%d", &n);
priority_queue<int> pq;
for (int i = 0; i < n; ++i) {
int x;
scanf("%d", &x);
pq.push(-x);
}
for (int i = 0; i < 1000; ++i) {
arr[i] = 0;
cnt[i] = 0;
}
while (!pq... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static boolean check(int[] arr, int no){
int[] piles = new int[no];
for(int i = 0; i < arr.length; i++){
boolean isPossible = false;
for(int j = 0; j < no; j++){
if(piles[j] <= arr[i]){
piles[j]++;
isPossible = tru... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, pile, cnt, ans = 0;
multiset<int> s;
multiset<int>::iterator it;
int main(void) {
ios ::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> pile;
s.insert(pile);
}
while (s.size()) {
s.erase(s.begin());
cnt = 1;... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long int inf = 1e18;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
vector<long long int> arr[100];
long long int n, t, i, p, j, cnt = 0;
cin >> n;
long long int v[n];
for (i = 0; i < n; i++) cin >> v[i];
sort(v, ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
const int inf = (int)1e9;
const int mod = 1e9 + 7;
using namespace std;
int a[111], ans = 0;
bool cmp(int x, int y) { return (x > y); }
bool u[111];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileWriter;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
import java.uti... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.*;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import java.util.Arrays;
import java.util.Collections;
import java.io.InputStream;
public class Main... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
vector<int> v;
vector<int> st;
vector<vector<int> > vv;
for (int i = 0; i < n; i++) {
int x;
scanf("%d", &x);
v.push_back(x);
}
sort(v.rbegin(), v.rend());
for (int i = 0; i < n; i++) {
int x = v[i];
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n;
vector<int> inp;
vector<vector<int> > dip;
void input() {
cin >> n;
inp.resize(n);
for (int i = 0; i < n; ++i) cin >> inp[i];
}
void owp() {
dip.resize(1);
dip[0].push_back(inp[0]);
for (int i = 1; i < n; ++i) {
int q = dip.size();
bool b = true;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[101];
bool vis[101] = {0};
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
int ans = 0, tot = n;
sort(a + 1, a + n + 1);
while (tot) {
int cnt = 0;
for (int i = 1; i <= n; i++) {
if (!vis[i] && cnt <= a[i... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
int num;
for (int i = 0; i < n; i++) {
cin >> num;
v.push_back(num);
}
sort(v.begin(), v.end());
int cnt = 0;
int all = 0;
vector<bool> taken(v.size(), false);
while (true) {
int cur = 0;
for... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, ans = 0, ps, j, f;
cin >> n;
int a[n], b[n], pile[n + 1];
for (i = 0; i < n; i++) {
cin >> a[i];
}
for (i = 0; i <= n; i++) {
pile[i] = 0;
}
sort(a, a + n);
ans = 1;
pile[ans]++;
for (i = 1; i < n; i++) {
f = 0;
f... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
@SuppressWarnings("unchecked")
public class P388A {
public void run() throws Exception {
TreeMap<Integer, Integer> xc = new TreeMap();
for (int n = nextInt(); n > 0; n--) {
int x = nextInt();
xc.put(x, xc.getOrD... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
long long int ans;
while (b > 0) {
if (b % 2 == 1) {
ans *= a;
}
b /= 2;
a *= a;
}
return ans;
}
struct comparator {
bool operator()(long long int i, long long int j) { return i > j; }
}... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n];
int c[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
c[i] = 0;
}
sort(a, a + n);
int w = 0;
;
for (int i = 0; i < n; i++) {
if (c[i] == 0) {
int x = c[i];
int e = 1;
w++;
c[i] = ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(input())
a = list(map(int, input().split()))
a.sort()
uni = {}
for elem in a:
if elem not in uni:
uni[elem]=1
else:
uni[elem]+=1
piles = []
while len(a)>0:
box = a.pop(0)
if len(piles)==0:
piles.append([box])
else:
for i in range(len(piles)):
temp... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int N;
vector<int> V;
bool solve(int n);
int main() {
ios_base::sync_with_stdio(false);
int i, st, dr, mid;
cin >> N;
V.resize(N);
for (i = 0; i < N; ++i) {
cin >> V[i];
}
sort(V.begin(), V.end());
st = 1;
dr = N;
while (st < dr) {
mid = (st + dr... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n;
int a[101], b[101];
int ans[101];
int main() {
int i, j;
while (cin >> n) {
memset(b, 0, sizeof(b));
for (i = 0; i < n; i++) cin >> a[i], b[a[i]]++;
memset(ans, 0, sizeof(ans));
int now = 0, Max = 0;
for (i = 0; i <= 100; i++) {
if (b[i]... | CPP |
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