Codeforces Round #586 (Div. 1 + Div. 2)

Codeforces Round #586 (Div. 1 + Div. 2)

A. Cards

• 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, cnt1, cnt2;
string s;
vector<int> cnt(26);

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 >> s;
    for (auto x: s)
        cnt[x - 'a'] ++ ;
    cnt1 = cnt['n' - 'a'];
    cnt2 = cnt['z' - 'a'];
    for (int i = 1; i <= cnt1; i ++ )
        cout << "1 ";
    for (int i = 1; i <= cnt2; i ++ )
        cout << "0 ";
    return 0;
}

B. Multiplication Table

• 思路：(frac{a_xa_y * a_xa_z}{a_ya_z} = {a_x}^2)

• 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 = 1e3 + 10;

int n;
ll ans[N];
ll M[N][N];

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;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            cin >> M[i][j];
    ans[1] = sqrt(M[1][2] * M[1][3] / M[2][3]);
    for (int i = 2; i <= n; i ++ )
        ans[i] = M[1][i] / ans[1];
    for (int i = 1; i < n; i ++ )
        cout << ans[i] << " ";
    cout << ans[n] << "
";
    return 0;
}

C. Substring Game in the Lesson

• 思路：瞎猜搞出来的

• 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;
}

char min_ = 'z';
string s;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> s;
    for (int i = 0; i < s.length(); i ++ ){
        if (min_ < s[i])
            cout << "Ann
";
        else{
            min_ = s[i];
            cout << "Mike
";
        }
    }
    return 0;
}

D. Alex and Julian

• 思路：参照dalao的思路 如果是二分图, 则不存在奇环,.
  考虑存在a,则有0->a, a->2a, 如果存在2a,则有0->2a,就存在奇环,如果有4a, 6a 也会存在奇环, 因为0->4a, 0->6a都属于同一边.
  考虑存在p, 有0->p, 同时可以存在, 3p, 5p, 考虑p的最高2的幂在p乘上一个奇数以后不会增加, 所以同时只能存在2的幂与p相等的值. 原文链接

• 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 = 70;

int n, m;
ll b[N], x;
vector<ll> a[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 >> n;
    for (int i = 1; i <= n; i ++ ){
        cin >> b[i];
        int tmp = 0;
        x = b[i];
        while (x % 2 == 0){
            x /= 2;
            tmp ++ ;
        }
        a[tmp].push_back(i);
    }
    m = -1;
    for (int i = 0; i <= 64; i ++ )
        if (m == -1 || a[m].size() < a[i].size())
            m = i;
    cout << n - a[m].size() << "
";
    for (int i = 0; i <= 64; i ++ )
        if (i != m)
            for (int j = 0; j < a[i].size(); j ++ )
                cout << b[a[i][j]] << " ";
    return 0;
}