AtCoder Beginner Contest 180

A - box

C++

#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<utility>
#include<iomanip>
#include<map>
#include<queue>
#include<cmath>
#include<cstdio>

#define rep(i,n) for(int i=0; i<(n); ++i)
#define pai 3.1415926535897932384

using namespace std;
using ll =long long;
using P = pair<int,int>;

int main(int argc, const char * argv[]) {
    int n, a, b;
    cin >> n >> a >> b;

    cout << n - a + b << endl;
    return 0;
}

B - Various distances

Python

import math
N = int(input())
X = list(map(int, input().split()))

ans1 = 0
ans2 = 0
ans3 = 0

for i in range(0, N):
    X[i] = abs(X[i]);
    ans1 = ans1 + X[i];
    ans2 = ans2 + X[i] * X[i];

print(ans1);
print(math.sqrt(ans2));
print(max(X));

C - Cream puff

C++

#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<utility>
#include<iomanip>
#include<map>
#include<queue>
#include<cmath>
#include<cstdio>

#define rep(i,n) for(int i=0; i<(n); ++i)
#define pai 3.1415926535897932384

using namespace std;
using ll =long long;
using P = pair<int,int>;

vector<ll> divisor(ll n) {
    vector<ll> ret;
    for (ll i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    sort(ret.begin(), ret.end());
    return ret;
}

int main(int argc, const char * argv[]) {
    ll N;
    cin >> N;

    vector<long long> Ans = divisor(N);
    rep(i, Ans.size()){
        cout << Ans[i] << endl;
    }

    return 0;
}

D - Takahashi Unevolved

C++

#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<utility>
#include<iomanip>
#include<map>
#include<queue>
#include<cmath>
#include<cstdio>

#define rep(i,n) for(int i=0; i<(n); ++i)
#define pai 3.1415926535897932384
#define ll_limit 2e18

using namespace std;
using ll =long long;
using P = pair<int,int>;

int main(int argc, const char * argv[]) {
    ll X, Y, A, B;
    cin >> X >> Y >> A >> B;
    ll ans = 0;

    while((double)A*X<=2e18 && A*X<=A+B && A*X<Y){
        X*=A;
        ans++;
    }

    cout << ans+(Y-1-X)/B << endl;
    return 0;
}