CARRYGOLD - Editorial

editorial
, ,
1

PROBLEM LINK:

Practice
Div1
Div2
Div3

Setter: Utkarsh Gupta
Tester: Samarth Gupta
Editorialist: Ajit Sharma Kasturi

DIFFICULTY:

CAKEWALK

PREREQUISITES:

None

PROBLEM:

There are a total of N+1 people visiting a gold mine. There is total of X kg of gold present and every person can carry atmost Y kg of gold with him. If each person can enter the goldmine exactly once, we need to find out whether those N+1 people can pick up the entire gold or not.

EXPLANATION:

  • Since every person can carry atmost Y kg of gold with him, N+1 persons can together carry atmost (N+1) \cdot Y kg of gold with them.

  • If this value is greater than or equal to X, we output YES else we output NO.

TIME COMPLEXITY:

O(N) for each testcase.

SOLUTION:

Editorialist's solution 
#include <bits/stdc++.h>
using namespace std;

int main()
{
     int tests;
     cin >> tests;
     while (tests--)
     {
          int n, x, y;
          cin >> n >> x >> y;
          if ((n + 1) * y >= x)
          {
               cout << "YES" << endl;
          }
          else
          {
               cout << "NO" << endl;
          }
     }
     return 0;
}
Setter's solution 
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val)  no. of elements strictly less than val
// s.find_by_order(i)  itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
void solve()
{
    ll n,x,y;
    cin>>n>>x>>y;
    assert(n>=1 && n<=1000);
    assert(x>=1 && x<=1000);
    assert(y>=1 && y<=1000);
    ll can=(n+1)*y;
    if(can>=x)
        cout<<"YES\n";
    else
        cout<<"NO\n";
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T=1;
    cin>>T;
    assert(T>=1 && T<=1000);
    int t=0;
    while(t++<T)
    {
        //cout<<"Case #"<<t<<":"<<' ';
        solve();
        //cout<<'\n';
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester's solution 
#include <bits/stdc++.h>
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,' ');
}
 
void readEOF(){
    assert(getchar()==EOF);
}

int main() {
	// your code goes here
	int t = readIntLn(1, 1000);
	int sum = 0;
	while(t--){
	    int n, x, y;
	    n = readIntSp(1, 1000);
	    x = readIntSp(1, 1000);
	    y = readIntLn(1, 1000);
	    cout << (x > y*(n + 1) ? "nO" : "YeS") << '\n';
	}

	return 0;
}

Please comment below if you have any questions, alternate solutions, or suggestions. :slight_smile: