Submission #1296422

#	Time	Username	Problem	Language	Result	Execution time	Memory
1296422		orzshadownova	A Huge Tower (CEOI10_tower)	C++20	0 / 100	1 ms	580 KiB

tower

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <iomanip>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <sstream>
#include <limits>
#include <functional>
// author: @satyam

using namespace std;

using i64 = long long;
using u64 = unsigned long long;
using ld  = long double;


using vi  = vector<int>;
using vi64 = vector<i64>;
using pii = pair<int,int>;
using pll = pair<i64,i64>;
typedef long long ll;

//#define int long long
#define pb push_back
#define ln "\n"
#define newline cout << ln
#define all(x) (x).begin(), (x).end()


//being stubborn is a good idea
/*
problem statement
point is a ,b,x is at max 10 u need to make sq>rec  8 8 2

*/

/*
if you are still not learning then nothing we can do u need to learn to apply it further



*/
/*
claims on algo

*/

int binarySearch(vector<int>& arr, int target) {
    int left = 0, right = arr.size() - 1;

    while (left <= right) {
        int mid = left + (right - left) / 2;

        if (arr[mid] == target) return mid;   // found
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1; // not found
}

//hashing and also how map is so f* cool
i64 gcdll (i64 a, i64 b) {
    if (b == 0)
        return a;
    else
        return gcdll (b, a % b);
}
long long power(long long base, long long exp, long long mod) {
    long long res = 1;
    base %= mod;
    while (exp > 0) {
        if (exp % 2 == 1) {
            res = (static_cast<__int128>(res) * base) % mod;
        }
        base = (static_cast<__int128>(base) * base) % mod;
        exp /= 2;
    }
    return res;
}
long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1)  // if b is odd
            res *= a;
        a *= a;
        b >>= 1;   // divide b by 2
    }
    return res;
}

long long modInverse(long long n, long long mod) {
    return power(n, mod - 2, mod);
}
int ceil_div(int a,int b) {
    return (a+b-1)/b;
}
/*
always gives u  two think and compared to have ratio its pretty simple if i have like ti as in w also atleats >1 u can
make matbe whats optimal  like ratio wld be t/w t-(ratio)*w >= 0 pllly bs bcs we need fir max
new state like that apply dp on w

*/



int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    // freopen("talent.in", "r", stdin);
    // freopen("talent.out", "w", stdout);
    int N, D;
    cin >> N >> D;
    i64 a , b;
    cin >> a >> b;
    if (a == b)cout << 0;
    else cout << 1 ;

}
/*
 its just step by step learn new thing then practice more harder
 */

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