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Practice
Setter: Daanish Mahajan
Tester & Editorialist: Taranpreet Singh
DIFFICULTY
Cakewalk
PREREQUISITES
None
PROBLEM
Ash is trying to get visa to Chefland. For the visa to be approved, he needs to satisfy the following three criteria:
- Solve at least x_1 problems on Codechef.
- Have at least y_1 current rating on Codechef.
- Make his last submission at most z_1 months ago.
You are given the number of problems solved by Chef (x_2), his current rating (y_2) and the information that he made his last submission z_2 months ago. Determine whether he will get the visa.
QUICK EXPLANATION
Ash can go to chefland if and only if x_2 \geq x_1, y_2 \geq y_1 and z_2 \leq z_1 holds.
EXPLANATION
This problem is just meant to test users familiarity with the programming language constructs, we just need to implement the mentioned conditions.
We need to check
- If the number of problems solved x_2 is at least x_1, implying we need to check whether x_2 \geq x_1 holds.
- If the current rating is at least y_2 is at least y_1, implying we need to check whether y_2 \geq y_1 holds.
- If the number of months since last submission z_2 is at most z_1, implying we need to check whether z_2 \leq z_1
If and only if all of the conditions are satisfied, Ash can go to Chefland.
One way to implement this would be using AND operator, denoted by && in most languages.
if x2 >= x1 && y2 >= y1 && z2 <= z1: YES
else: NO
Another way without AND operator is as follows, by making a count variable, storing the number of conditions satisfied.
count = 0
if x2 >= x1: count++
if y2 >= y1: count++
if z2 <= z1: count++
if count == 3: YES
else: NO
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;
const int maxt = 5000;
const string newln = "\n", space = " ";
int main()
{
int t, x, x1, y, y1, z, z1; cin >> t;
while(t--){
cin >> x >> x1 >> y >> y1 >> z >> z1;
string ans = (x1 >= x && y1 >= y && z1 <= z) ? "YeS" : "No";
cout << ans << endl;
}
}
Tester's Solution
#include <iostream>
#include <assert.h>
#include <algorithm>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <map>
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){
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 T = readIntLn(1, 5000);
while(T-->0){
int x1 = readIntSp(20, 50), x2 = readIntSp(20, 50);
int y1 = readIntSp(1900, 2100), y2 = readIntSp(1900, 2100);
int z1 = readIntSp(1, 6), z2 = readIntLn(1, 6);
int count = 0;
if(x2 >= x1)count++;
if(y2 >= y1)count++;
if(z2 <= z1)count++;
if(count == 3)cout<<"YES\n";
else cout<<"NO\n";
}
assert(getchar()==-1);
cerr<<"SUCCESS\n";
}
Feel free to share your approach. Suggestions are welcomed as always. 