🚀 16-Aug-2021

Author: sureshmangs

Diff for: competitive programming/codechef/COPR16G.cpp

@@ -0,0 +1,56 @@
+/*
+The easiest problem of the lot !
+
+You are provided with coins of denominations "a" and "b". You are required to find the least value of n, such that all currency values greater than or equal to n can be made using any number of coins of denomination "a" and "b" and in any order. If no such number exists, print "-1" in that case.
+
+Input Constraints:
+1 <= t <= 106
+1 <= a <= 109
+1 <= b <= 109
+Input Format:
+The first line contains t, the number of test cases.
+
+Each of the next t lines contains 2 integers a & b, as specified in the question.
+
+Output Format:
+For each test case, print the required answer
+
+Example
+Input:
+1 
+4 7
+6 1
+2 4
+
+Output:
+18
+0
+-1
+NOTE:Prefer fast input/output methods
+*/
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t;
+	cin >> t;
+	
+	while (t--) {
+		long long a, b;
+		cin >> a >> b;
+		
+		int g = __gcd(a, b);
+		
+		if (g > 1) cout << -1 << "\n";
+		else cout << a * b - a - b + 1 << "\n"; // using chicken mcnugget theorem
+	}
+	
+    return 0;
+}

Diff for: competitive programming/codeforces/1182A - Filling Shapes.cpp

@@ -0,0 +1,53 @@
+/*
+You have a given integer n. Find the number of ways to fill all 3×n tiles with the shape described in the picture below. Upon filling, no empty spaces are allowed. Shapes cannot overlap.
+
+This picture describes when n=4. The left one is the shape and the right one is 3×n tiles.
+Input
+The only line contains one integer n (1=n=60) — the length.
+
+Output
+Print the number of ways to fill.
+
+Examples
+inputCopy
+4
+outputCopy
+4
+inputCopy
+1
+outputCopy
+0
+Note
+In the first example, there are 4 possible cases of filling.
+
+In the second example, you cannot fill the shapes in 3×1 tiles
+*/
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+void solve() {
+	int n;
+	cin >> n;
+	
+	if (n % 2) cout << 0 << "\n";
+	else cout << (int)pow(2, n / 2) << "\n";
+}
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t = 1;
+	//cin >> t;
+	
+	while (t--)  solve();
+	
+    return 0;
+}

Diff for: competitive programming/codeforces/1555C - Coin Rows.cpp

@@ -0,0 +1,166 @@
+Alice and Bob are playing a game on a matrix, consisting of 2 rows and m columns. The cell in the i-th row in the j-th column contains ai,j coins in it.
+
+Initially, both Alice and Bob are standing in a cell (1,1). They are going to perform a sequence of moves to reach a cell (2,m).
+
+The possible moves are:
+
+Move right — from some cell (x,y) to (x,y+1);
+Move down — from some cell (x,y) to (x+1,y).
+First, Alice makes all her moves until she reaches (2,m). She collects the coins in all cells she visit (including the starting cell).
+
+When Alice finishes, Bob starts his journey. He also performs the moves to reach (2,m) and collects the coins in all cells that he visited, but Alice didn't.
+
+The score of the game is the total number of coins Bob collects.
+
+Alice wants to minimize the score. Bob wants to maximize the score. What will the score of the game be if both players play optimally?
+
+Input
+The first line contains a single integer t (1=t=104) — the number of testcases.
+
+Then the descriptions of t testcases follow.
+
+The first line of the testcase contains a single integer m (1=m=105) — the number of columns of the matrix.
+
+The i-th of the next 2 lines contain m integers ai,1,ai,2,…,ai,m (1=ai,j=104) — the number of coins in the cell in the i-th row in the j-th column of the matrix.
+
+The sum of m over all testcases doesn't exceed 105.
+
+Output
+For each testcase print a single integer — the score of the game if both players play optimally.
+
+Example
+inputCopy
+3
+3
+1 3 7
+3 5 1
+3
+1 3 9
+3 5 1
+1
+4
+7
+outputCopy
+7
+8
+0
+Note
+The paths for the testcases are shown on the following pictures. Alice's path is depicted in red and Bob's path is depicted in blue.
+
+
+
+
+
+
+
+// o(m) accepted
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+void solve() {
+	int m;
+	cin >> m;
+	
+	vector <vector<int>> v(2, vector<int> (m));
+	
+	for (int i = 0; i < 2; i++) {
+		for (int j = 0; j <  m; j++) {
+			cin >> v[i][j];
+		}
+	}
+	
+	vector <vector<int>> psum(2, vector<int> (m));
+	
+	psum[0][0] = v[0][0];
+	psum[1][0] = v[1][0];
+	
+	for (int i = 0; i < 2; i++) {
+		for (int j = 1; j <  m; j++) {
+			psum[i][j] = psum[i][j - 1] + v[i][j];
+		}
+	}
+	
+	int res = INT_MAX;
+	
+	for (int i = 0; i < m; i++) {
+		// 0th row
+		int up = psum[0][m - 1] - psum[0][i];
+		
+		
+		// 1st row
+		int down = 0;
+		
+		if (i - 1 >= 0) down = psum[1][i - 1];
+		
+		res = min(res, max(up, down));
+	}
+	
+	cout << res << "\n";
+}
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t = 1;
+	cin >> t;
+	
+	while (t--)  solve();
+	
+    return 0;
+}
+
+
+
+
+
+
+
+// o(m^2) TLE
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+void solve() {
+	int m;
+	cin >> m;
+	
+	vector <vector<int>> v(2, vector<int> (m));
+	
+	for (int i = 0; i < 2; i++) {
+		for (int j = 0; j <  m; j++) {
+			cin >> v[i][j];
+		}
+	}
+	
+	int res = INT_MAX;
+	
+	for (int i = 0; i < m; i++) {
+		int up = 0;
+		
+		// 0th row
+		for (int j = i + 1; j < m; j++) up += v[0][j];
+		
+		int down = 0;
+		
+		// 1st row
+		for (int j = 0; j <  i; j++) down += v[1][j];
+		
+		res = min(res, max(up, down));
+	}
+	
+	cout << res << "\n";
+}
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t = 1;
+	cin >> t;
+	
+	while (t--)  solve();
+	
+    return 0;
+}

Diff for: competitive programming/codeforces/1557A - Ezzat and Two Subsequences.cpp

@@ -0,0 +1,96 @@
+/*
+Ezzat has an array of n integers (maybe negative). He wants to split it into two non-empty subsequences a and b, such that every element from the array belongs to exactly one subsequence, and the value of f(a)+f(b) is the maximum possible value, where f(x) is the average of the subsequence x.
+
+A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.
+
+The average of a subsequence is the sum of the numbers of this subsequence divided by the size of the subsequence.
+
+For example, the average of [1,5,6] is (1+5+6)/3=12/3=4, so f([1,5,6])=4.
+
+Input
+The first line contains a single integer t (1=t=103)— the number of test cases. Each test case consists of two lines.
+
+The first line contains a single integer n (2=n=105).
+
+The second line contains n integers a1,a2,…,an (-109=ai=109).
+
+It is guaranteed that the sum of n over all test cases does not exceed 3·105.
+
+Output
+For each test case, print a single value — the maximum value that Ezzat can achieve.
+
+Your answer is considered correct if its absolute or relative error does not exceed 10-6.
+
+Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if |a-b|max(1,|b|)=10-6.
+
+Example
+inputCopy
+4
+3
+3 1 2
+3
+-7 -6 -6
+3
+2 2 2
+4
+17 3 5 -3
+outputCopy
+4.500000000
+-12.500000000
+4.000000000
+18.666666667
+Note
+In the first test case, the array is [3,1,2]. These are all the possible ways to split this array:
+
+a=[3], b=[1,2], so the value of f(a)+f(b)=3+1.5=4.5.
+a=[3,1], b=[2], so the value of f(a)+f(b)=2+2=4.
+a=[3,2], b=[1], so the value of f(a)+f(b)=2.5+1=3.5.
+Therefore, the maximum possible value 4.5.
+In the second test case, the array is [-7,-6,-6]. These are all the possible ways to split this array:
+
+a=[-7], b=[-6,-6], so the value of f(a)+f(b)=(-7)+(-6)=-13.
+a=[-7,-6], b=[-6], so the value of f(a)+f(b)=(-6.5)+(-6)=-12.5.
+Therefore, the maximum possible value -12.5.
+*/
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+void solve() {
+	long long n;
+	cin >> n;
+	
+	vector <long long> v(n);
+	
+	for (auto &x: v) cin >> x;
+	
+	int maxi = *max_element(v.begin(), v.end());
+	
+	long long sum = 0;
+	
+	for (auto &x: v) sum += x;
+	
+	sum -= maxi;
+	
+	double res = double(maxi) + double(sum) / (n - 1);
+	
+	cout << setprecision(9) << fixed << res << "\n";
+	
+}
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t = 1;
+	cin >> t;
+	
+	while (t--)  solve();
+	
+    return 0;
+}