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Practice
Setter: Daanish Mahajan
Tester: Istvan Nagy
Editorialist: Taranpreet Singh
DIFFICULTY
Cakewalk
PREREQUISITES
None
PROBLEM
Chefland has 2 different types of coconut, type A and type B. Type A contains only x_a milliliters of coconut water and type B contains only x_b grams of coconut pulp. Chef’s nutritionist has advised him to consume X_a milliliters of coconut water and X_b grams of coconut pulp every week in the summer. Find the total number of coconuts (type A + type B) that the Chef should buy each week to keep himself active in the hot weather.
QUICK EXPLANATION
Chef must eat X_a/x_a coconuts of type A and X_b/x_b coconuts of type B.
EXPLANATION
Chef gets x_a milliliters of coconut water from coconuts of type A, and he needs X_a milliliters of coconut water, then how many coconuts of type A does the chef need?
Since 1 coconut provides x_a milliliters of coconut water, P coconuts would provide P*x_a milliliters. Chef need P such that P*x_a = X_a, which implies P = X_a / x_a. Hence, Chef consume P coconuts of type A.
By similar reasoning, we can see that Chef consumes Q = X_b/x_b coconuts of type B.
This way, the total number of coconuts consumed is P+Q = X_a/x_a + X_b/x_b.
Following C++ code depicts the above idea
int T;
cin>>T;
while(T-->0){
int Xa, xa, Xb, xb;
cin>>xa>>xb>>Xa>>Xb;
int count = Xa/xa + Xb/xb;
cout<<count<<"\n";
}
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
#include<bits/stdc++.h>
# define pb push_back
#define pii pair<int, int>
#define mp make_pair
# define ll long long int
using namespace std;
int main()
{
int t; cin >> t;
int a, b, c, d;
while(t--){
cin >> a >> b >> c >> d;
int ans = c / a + d / b;
cout << ans << endl;
}
}
Tester's Solution
#include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>
#ifdef HOME
#include <windows.h>
#endif
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
template<class T> bool umin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T &a, T b) { return a < b ? (a = b, true) : false; }
using namespace std;
long long readInt(long long l, long long r, char endd) {
long long x = 0;
int cnt = 0;
int fi = -1;
bool is_neg = false;
while (true) {
char g = getchar();
if (g == '-') {
assert(fi == -1);
is_neg = true;
continue;
}
if ('0' <= g && g <= '9') {
x *= 10;
x += g - '0';
if (cnt == 0) {
fi = g - '0';
}
cnt++;
assert(fi != 0 || cnt == 1);
assert(fi != 0 || is_neg == false);
assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
}
else if (g == endd) {
assert(cnt > 0);
if (is_neg) {
x = -x;
}
assert(l <= x && x <= r);
return x;
}
else {
//assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret = "";
int cnt = 0;
while (true) {
char g = getchar();
assert(g != -1);
if (g == endd) {
break;
}
cnt++;
ret += g;
}
assert(l <= cnt && cnt <= r);
return ret;
}
long long readIntSp(long long l, long long r) {
return readInt(l, r, ' ');
}
long long readIntLn(long long l, long long r) {
return readInt(l, r, '\n');
}
string readStringLn(int l, int r) {
return readString(l, r, '\n');
}
string readStringSp(int l, int r) {
return readString(l, r, ' ');
}
int main(int argc, char** argv)
{
#ifdef HOME
if(IsDebuggerPresent())
{
freopen("../in.txt", "rb", stdin);
freopen("../out.txt", "wb", stdout);
}
#endif
int T = readIntLn(1, 15'000);
forn(tc, T)
{
int xa = readIntSp(100, 200);
int xb = readIntSp(400, 500);
int Xa = readIntSp(1'000, 1'200);
int Xb = readIntLn(1'000, 1'500);
assert(Xa % xa == 0);
assert(Xb % xb == 0);
printf("%d\n", Xa / xa + Xb / xb);
}
assert(getchar() == -1);
return 0;
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class COCONUT{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int xa = ni(), xb = ni(), Xa = ni(), Xb = ni();
pn(Xa/xa + Xb/xb);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new COCONUT().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
Feel free to share your approach. Suggestions are welcomed as always. 