AtCoder Beginner Contest 144

AtCoder Beginner Contest 144

A - 9x9

• 思路：水题

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int a, b;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> a >> b;
    if (a <= 9 && b <= 9)
        cout << a * b << "
";
    else
        cout << "-1
";
    return 0;
}

B - 81

• 思路：水题

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int n;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    bool flag = false;
    for (int a = 1; a <= 9; a ++ ){
        for (int b = 1; b <= 9; b ++ ){
            if (a * b == n){
                flag = true;
                break;
            }
        }
        if (flag)
            break;
    }
    if (flag)
        cout << "Yes
";
    else
        cout << "No
";
    return 0;
}

C - Walk on Multiplication Table

• 思路：暴力即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

ll n, tmp, ans;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    tmp = sqrt(n);
    for (ll i = tmp; i >= 1; i -- ){
        if (n % i == 0){
            ans = i + n / i - 2;
            break;
        }
    }
    cout << ans << "
";
    return 0;
}

D - Water Bottle

• 思路：两种情况比较一下即可

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const double pi = acos(-1);

double a, b, x, s, h, ans;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> a >> b >> x;
    s = x / a;
    if (s >= a * b / 2){
        h = (a * b - s) * 2 / a;
        ans = 180 * atan2(h, a) / pi;
    }
    else{
        h = s * 2 / b;
        ans = 180 * atan2(b, h) / pi;
    }
    printf("%.8lf
", ans);
    return 0;
}

E - Gluttony

• 思路：二分 上界为最大的最小(a)(*)最大(f)

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 2e5 + 10;

ll n, k, x, l, r, mid, tmp;
vector<ll> a, f;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ ){
        cin >> x;
        a.push_back(x);
    }
    for (int i = 1; i <= n; i ++ ){
        cin >> x;
        f.push_back(x);
    }
    sort(a.begin(), a.end());
    sort(f.rbegin(), f.rend());
    for (int i = 0; i < n; i ++ )
        r = max(r, a[i] * f[i]);
    while (l < r){
        mid = (l + r) >> 1;
        tmp = 0;
        for (int i = 0; i < n; i ++ )
            tmp += max(0ll, a[i] - mid / f[i]);
        if (tmp <= k)
            r = mid;
        else
            l = mid + 1;
    }
    cout << l << "
";
    return 0;
}