Codeforces Round #587 (Div. 3)

Posted 2020-11-30 misuchii

Codeforces Round #587 (Div. 3)

A. Prefixes

• 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, cnt;
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; i += 2){
        if (s[i] == s[i + 1]){
            cnt ++ ;
            if (s[i] == 'b')
                s[i] = 'a';
            else
                s[i] = 'b';
        }
    }
    cout << cnt << "
" << s << "
";
    return 0;
}

B. Shooting

• 思路：贪心 写个struct 从大到小排序计算

• 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 = 1010;

struct node{
    int a, id;
}a[N];

inline bool cmp(const node &a, const node &b){
    return a.a > b.a;
}

int n, 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;
    for (int i = 1; i <= n; i ++ ){
        cin >> a[i].a;
        a[i].id = i;
    }
    sort(a + 1, a + n + 1, cmp);
    for (int i = 1; i <= n; i ++ )
        ans += a[i].a * (i - 1) + 1;
    cout << ans << "
";
    for (int i = 1; i < n; i ++ )
        cout << a[i].id << " ";
    cout << a[n].id << "
";
    return 0;
}

C. White Sheet

• 思路：把所有不可能的情况枚举出来

• 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 x1, y1_, x2, y2, x3, y3, x4, y4, x5, y5, x6, y6;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> x1 >> y1_ >> x2 >> y2 >> x3 >> y3 >> x4 >> y4 >> x5 >> y5 >> x6 >> y6;
    if ((x1 >= x3 && x2 <= x4 && y1_ >= y3 && y2 <= y4) || (x1 >= x5 && x2 <= x6 && y1_ >= y5 && y2 <= y6))
        cout << "NO
";
    else if ((y1_ >= y3 && y2 <= y4 && y1_ >= y5 && y2 <= y6) && (x1 >= x3 && x2 <= x6 && x4 >= x5 || x1 >= x5 && x2 <= x4 && x3 <= x6))
        cout << "NO
";
    else if ((x1 >= x3 && x2 <= x4 && x1 >= x5 && x2 <= x6) && (y1_ >= y3 && y2 <= y6 && y4 >= y5 || y1_ >= y5 && y2 <= y4 && y3 <= y6))
        cout << "NO
";
    else
        cout << "YES
";
    return 0;
}

D. Swords

• 思路：排序后求最大的和其他每一项的差之和与gcd 答案为(sum / gcd)和(gcd)

• 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;
ll sum, gcd_;
ll a[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 ++ )
        cin >> a[i];
    sort(a + 1, a + n + 1);
    for (int i = 1; i < n; i ++ ){
        sum += a[n] - a[i];
        gcd_ = gcd(gcd_, a[n] - a[i]);
    }
    cout << sum / gcd_ << " " << gcd_ << "
";
    return 0;
}

E1. Numerical Sequence (easy version)

• 思路：只会easy version 暴力搞

• 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 q, tot;
ll k;
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 >> q;
    while (q -- ){
        tot = 1;
        s.clear();
        cin >> k;
        while (s.length() < k){
            k -= s.length();
            s += to_string(tot ++ );
        }
        cout << s[k - 1] << "
";
    }
    return 0;
}