Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2)

Posted 2021-03-24 misuchii

Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2)

A. Forgetting Things

• 思路：(a)和(b)相等或者(a + 1 = b)或者(a = 9) (b = 1)才可

• 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;
}

int a, b, x;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> a >> b;
    if (a == 9 && b == 1){
        cout << "9 10
";
        return 0;
    }
    if (a == b)
        cout << a << "0 " << b << "1
";
    else if (a + 1 == b)
        cout << a << " " << b << "
";
    else
        cout << -1 << "
";
    return 0;
}

B. TV Subscriptions

• 思路：滑动窗口 辣鸡评测鸡 昨晚上交system test给我t了22 结果我早上交相同的代码就过了

• 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;
const int M = 1e6 + 10;
const int INF = 0x3f3f3f3f;

int t, n, k, d, min_, ans;
int a[N], b[M];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while(t -- ){
        memset(b, 0, sizeof(b));
        cin >> n >> k >> d;
        ans = 0;
        for (int i = 1; i <= n; i ++ )
            cin >> a[i];
        for (int i = 1; i <= d; i ++ )
            if (b[a[i]] ++ == 0)
                ans ++ ;
        min_ = ans;
        for (int i = d + 1; i <= n; i ++ ){
            if (( -- b[a[i - d]]) == 0)
                ans -- ;
            if ((b[a[i]] ++ ) == 0)
                ans ++ ;
            if (ans < min_)
                min_ = ans;
        }
        cout << min_ << "
";
    }
    return 0;
}

C. p-binary

• 思路：从1开始枚举ans 统计(n - ans * p)二进制中一的个数即可
  昨晚上没想到_builtin_popcountll 睡醒早上才想到
• 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, p, ans = 1;
 
int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> p;
    while (n - ans * p > 0){
        if (__builtin_popcountll(n - ans * p) <= ans){
            if (n - ans * p >= ans){
                cout << ans << "
";
                return 0;
            }
        }
        ans ++ ;
    }
    cout << "-1
";
    return 0;
}

D. Power Products

• 思路：照着官方题解写的 先将(a)唯一分解定理 对于每一项(bi ^ ki) 当且仅当(ki % k == 0)时才对结果有贡献 最后统计下即可

• 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(ll a, ll b){
    ll res = 1;
    while (b){
        if (b & 1)
            res *= a;
        a *= a;
        b >>= 1;
    }
    return res;
}

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 = 1e5 + 10;

int n, k, a;
ll ans;
map<vector<pair<int, int> >, int > mp;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ ){
        vector<pair<int, int> > vec, res;
        cin >> a;
        for (int b = 2; b * b <= a; b ++ ){
            int k_ = 0;
            while (a % b == 0){
                k_ ++ ;
                a /= b;
            }
            if (k_ % k)
                vec.push_back(make_pair(b, k_ % k));
        }
        if (a > 1)
            vec.push_back(make_pair(a, 1));
        for (int j = 0; j < vec.size(); j ++ )
            res.push_back(make_pair(vec[j].first, k - vec[j].second));
        ans += mp[res];
        mp[vec] ++ ;
    }
    cout << ans << "
";
    return 0;
}