Codeforces Round #585 (Div. 2)

Codeforces Round #585 (Div. 2)

A. Yellow 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 a1, a2, k1, k2, n, sum, res, cnt1, cnt2;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> a1 >> a2 >> k1 >> k2 >> n;
    if (k1 > k2){
        swap(a1, a2);
        swap(k1, k2);
    }
    sum = a1 * (k1 - 1) + a2 * (k2 - 1);
    res = max(0, min(a1 + a2, n - sum));
    cnt1 = min(a1, n / k1);
    n -= cnt1 * k1;
    cnt2 = min(a2, n / k2);
    cout << res << " " << cnt1 + cnt2 << "\n";
    return 0;
}

B. The Number of Products

• 思路：将正数令为0 负数令为1 统计前缀和 然后从头扫一遍 维护前缀和为奇数的个数和前缀和为偶数的个数

• 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;

int n;
int a[N], sum[N];
ll ans1, ans2, cnt1, cnt2;

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 >> a[i];
        if (a[i] > 0)
            a[i] = 0;
        else
            a[i] = 1;
    }
    for (int i = 1; i <= n; i ++ )
        sum[i] = (sum[i - 1] + a[i]) % 2;
    cnt1 ++ ;
    for (int i = 1; i <= n; i ++ ){
        if (sum[i]) {
            ans1 += cnt2;
            ans2 += cnt1;
            cnt2 ++ ;
        }
        else{
            ans1 += cnt1;
            ans2 += cnt2;
            cnt1 ++ ;
        }
    }
    cout << ans2 << " " << ans1 << "\n";
    return 0;
}

C. Swap Letters

• 思路：首先a和b都为奇数时不能变为相同的 然后两个ab或ba可以一步变成相同的 ab和ba要两步

• 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, cnta, cntb;
string s, t;
vector<int> pos_ab, pos_ba;
vector<pair<int, int> > ans;

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 >> t;
    for (int i = 0; i < n; i ++ ){
        if (s[i] == 'a')
            cnta ++ ;
        else
            cntb ++ ;
        if (t[i] == 'a')
            cnta ++ ;
        else
            cntb ++ ;
        if (s[i] != t[i]){
            if (s[i] == 'a')
                pos_ab.push_back(i + 1);
            else
                pos_ba.push_back(i + 1);
        }
    }
    if ((cnta & 1) || (cntb & 1)){
        cout << "-1\n";
        return 0;
    }
    for (int i = 0; i + 1 < pos_ab.size(); i += 2)
        ans.push_back(make_pair(pos_ab[i], pos_ab[i + 1]));
    for (int i = 0; i + 1 < pos_ba.size(); i += 2)
        ans.push_back(make_pair(pos_ba[i], pos_ba[i + 1]));
    if (pos_ab.size() & 1){
        ans.push_back(make_pair(pos_ab.back(), pos_ab.back()));
        ans.push_back(make_pair(pos_ab.back(), pos_ba.back()));
    }
    cout << ans.size() << "\n";
    for (auto it: ans)
        cout << it.first << " " << it.second << "\n";
    return 0;
}

D. Ticket Game

• 思路：博弈论

• 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, sum1, sum2, c1, c2;
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 >> n >> s;
    for (int i = 0; i < n / 2; i ++ ){
        if (s[i] == '?')
            c1 ++ ;
        else
            sum1 += s[i] - '0';
    }
    for (int i = n / 2; i < n; i ++ ){
        if (s[i] == '?')
            c2 ++ ;
        else
            sum2 += s[i] - '0';
    }
    if (!c1 && !c2){
        if (sum1 == sum2)
            cout << "Bicarp\n";
        else
            cout << "Monocarp\n";
    }
    else if (c1 == c2){
        if (abs(sum1 - sum2) >= 1)
            cout << "Monocarp\n";
        else
            cout << "Bicarp\n";
    }
    else if (c1 > c2){
        if (sum2 - sum1 == 9 * (c1 - c2) / 2)
            cout << "Bicarp\n";
        else
            cout << "Monocarp\n";
    }
    else if (c1 < c2){
        if (sum1 - sum2 == 9 * (c2 - c1) / 2)
            cout << "Bicarp\n";
        else
            cout << "Monocarp\n";
    }
    return 0;
}