CSES Problemset

cses/README.md

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+CSES is one of the most complete problemsets on the internet, with over 300 problems of all topics.
+https://cses.fi/problemset/list/
+
+In here we have all solutions to the problems of the dp and graph sections
+
+All the problems can be tested on the cses website, where you can also access the test cases, and once you solved the problem, access other solutions

cses/dp/1 - Dice Combinations.cpp

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+// Dice Combinations
+//
+// Problem name: Dice Combinations
+// Problem Link: https://cses.fi/problemset/task/1633
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int mod = 1e9 + 7;
+int dp[1000005];
+
+int dice(int n) {
+    int resp = 0;
+
+    // caso base
+    if (n == 0) return 1;
+
+    // se ja calculamos a dp
+    if (dp[n] > 0) {
+        return dp[n];
+    }
+    for (int i = 1; i<= 6; i++) {
+        if (n - i >= 0) {
+            resp += dice(n-i);
+
+            // lembrar de tirar o modulo
+            resp %= mod;
+        }
+    }
+    return dp[n] = resp;
+}
+
+int main () {
+    int n;
+    cin >> n;
+    cout << dice(n) << endl;
+    return 0;
+}

cses/dp/10 - Edit Distance.cpp

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+// Edit Distance
+//
+// Problem name: Edit Distance
+// Problem Link: https://cses.fi/problemset/task/1639
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+// The edit distance between two strings is the minimum number of operations required to transform one string into the other.
+// The allowed operations are:
+//     Add one character to the string.
+//     Remove one character from the string.
+//     Replace one character in the string.
+// For example, the edit distance between LOVE and MOVIE is 2, because you can first replace L with M, and then add I.
+// Your task is to calculate the edit distance between two strings.
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    string a, b;
+    cin >> a >> b;
+    int tama = (int)a.size(), tamb = (int)b.size();
+    vector<vector<int>> dp(5050, vector<int> (5050, 10000));
+    dp[tama][tamb] = 0;
+    for (int i = tama; i >= 0; i--) {
+        for (int j = tamb; j >= 0; j--) {
+            dp[i][j] = min(dp[i][j], (a[i] != b[j]) + dp[i+1][j+1]);
+            dp[i][j] = min(dp[i][j], 1 + dp[i+1][j]);
+            dp[i][j] = min(dp[i][j], 1 + dp[i][j+1]);
+        }
+    }
+    cout << dp[0][0] << '\n';
+    return 0;
+}

cses/dp/11 - Rectangle Cutting.cpp

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+// Rectangle Cutting
+//
+// Problem name: Rectangle Cutting
+// Problem Link: https://cses.fi/problemset/task/1744
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+// Given an a×b rectangle, your task is to cut it into squares. 
+// On each move you can select a rectangle and cut it into two rectangles 
+// in such a way that all side lengths remain integers. What is the minimum possible number of moves?
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    int a, b;
+    cin >> a >> b;
+    vector<vector<int>> dp(505, vector<int> (505, INF));
+    for (int i = 1; i <= a; i++) {
+        dp[i][i] = 0;
+        for (int j = 1; j <= b; j++) {
+            for (int k = 1; k < j; k++) {
+                dp[i][j] = min(dp[i][j], 1 + dp[i][j-k] + dp[i][k]);
+            }
+            for (int k = 1; k < i; k++) {
+                dp[i][j] = min(dp[i][j], 1 + dp[i-k][j] + dp[k][j]);
+            }
+        }
+    }
+    cout << dp[a][b] << '\n';
+    return 0;
+}

cses/dp/12 - Money Sums.cpp

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+// Money Sums
+//
+// Problem name: Money Sums
+// Problem Link: https://cses.fi/problemset/task/1745
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    int n;
+    cin >> n;
+    vector<int> v(n);
+    for (int i = 0; i < n; i++) cin >> v[i];
+    vector<int> dp(1e5 + 5);
+    dp[0] = 1;
+    int c = 0;
+    for (int i = 0; i < n; i++) {
+        for (int j = 1e5; j >= 0; j--) {
+            if (dp[j] && j+v[i] <= 1e5) {
+                c += (dp[j+v[i]] == 0);
+                dp[j+v[i]] = 1;
+            }
+        }
+    }
+    cout << c << '\n';
+    for (int i = 1; i <= 1e5; i++) 
+        if (dp[i]) cout << i << ' ';
+    cout << '\n';
+    return 0;
+}

cses/dp/13 - Removal game.cpp

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+// Removal Game
+//
+// Problem name: Removal Game
+// Problem Link: https://cses.fi/problemset/task/1097
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    int n;
+    cin >> n;
+    vector<int> v(n);
+    vector<vector<ll>> dp(n, vector<ll> (n));
+    ll soma = 0;
+    for (int i = 0; i < n; i++) {
+        cin >> v[i];
+        soma += v[i];
+        dp[i][i] = v[i];
+    }
+    for (int j = 1; j < n; j++) {
+        for (int i = 0; i < n; i++) {
+            if (i + j < n) {
+                dp[i][i+j] = max(v[i] - dp[i+1][i+j], v[i+j] - dp[i][i+j-1]);
+            }
+        }
+    }
+    cout << (soma+dp[0][n-1])/2 << '\n';
+    return 0;
+}

cses/dp/14 - Two Sets II.cpp

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+// Two Sets II
+//
+// Problem name: Two Sets II
+// Problem Link: https://cses.fi/problemset/task/1093
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int fexp(ll b, int e) {
+    ll resp = 1;
+    while (e) {
+        if (e&1) resp = (resp * b) % M;
+        b = (b * b) % M;
+        e = (e >> 1);
+    }
+    return resp;
+}
+
+int main () { _
+    int n;
+    cin >> n;
+    if ((n%4) && ((n+1)%4)) cout << "0\n";
+    else {
+        int meta = n*(n+1)/4;
+        vector<ll> dp(meta+1);
+        dp[0] = 1;
+        for (int i = 1; i <= n; i++) {
+            for (int j = meta; j >= 0; j--) {
+                if (j+i <= meta) {
+                    dp[j+i] += dp[j];
+                    if (dp[j+i] >= M) dp[j+i] -= M;
+                }
+            }
+        }
+        cout << (dp[meta] * fexp(2, M-2)) % M << '\n';
+    }
+    return 0;
+}

cses/dp/15 - Projects.cpp

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+// Projects
+//
+// Problem name: Projects
+// Problem Link: https://cses.fi/problemset/task/1140
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    int n;
+    cin >> n;
+    vector<pair<pii, int>> v(n);
+    vector<int> c;
+    for (int i = 0; i < n; i++) {
+        cin >> v[i].F.F >> v[i].F.S >> v[i].S;
+        c.pb(v[i].F.F), c.pb(v[i].F.S);
+    }
+    sort(c.begin(), c.end());
+    map<int, int> mapa;
+    int atual = 0;
+    mapa[c[0]] = atual++;
+    for (int i = 1; i < (int)c.size(); i++) {
+        if (c[i] != c[i-1]) mapa[c[i]] = atual++;
+    }
+    for (int i = 0; i < n; i++) {
+        v[i].F.F = mapa[v[i].F.F];
+        v[i].F.S = mapa[v[i].F.S];
+    }
+    sort(v.begin(), v.end());
+    vector<ll> dp(1e6);
+    atual = n-1;
+    for (int i = 5e5; i >= 0; i--) {
+        dp[i] = max(dp[i], dp[i+1]);
+        while (atual >= 0 && i == v[atual].F.F) {
+            dp[i] = max(dp[i], v[atual].S + dp[v[atual].F.S + 1]);
+            atual--;
+        }
+    }
+    cout << dp[0] << '\n';
+    return 0;
+}

cses/dp/16 - Increasing Subsequence.cpp

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+// Increasing Subsequence
+//
+// Problem name: Increasing Subsequence
+// Problem Link: https://cses.fi/problemset/task/1145
+// Author: Bernardo Archegas (https://codeforces.com/profile/Ber)
+
+// You are given an array containing n integers. Your task is to determine the longest 
+//increasing subsequence in the array, i.e., 
+// the longest subsequence where every element is larger than the previous one.
+// A subsequence is a sequence that can be derived from the array by deleting 
+// some elements without changing the order of the remaining elements. 
+
+#include <bits/stdc++.h>
+#define _ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+#define MAXN 1000100
+#define INF 100000000
+#define pb push_back
+#define F first
+#define S second
+
+using namespace std;
+typedef long long int ll;
+typedef pair<int, int> pii;
+const int M = 1e9+7;
+
+int main () { _
+    int n;
+    cin >> n;
+    vector<int> v(n), pilha;
+    for (int i = 0; i < n; i++) {
+        cin >> v[i];
+        auto it = lower_bound(pilha.begin(), pilha.end(), v[i]);
+        if (it == pilha.end()) {
+            pilha.pb(v[i]);
+        }
+        else *it = v[i];
+    }
+    cout << (int)pilha.size() << '\n';
+    return 0;
+}