Codeforces Round #651 (Div. 2)

这几天小学期, 要肝程序, 晚上tm还上物理实验课就离谱, 和cf冲突, 游戏还有活动~
这几天cf只能随缘补题, 打比赛估计不可能, 和物理实验冲突
好亏啊,这场能上分

A

氵题

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n;
        if (n <= 3) cout << 1 << '\n';
        else cout << (n >> 1) << '\n';
    }
    return 0;
}

B

看清题奥, 又公因数就行, 直接gcd == 2多简单, 最后的扔掉就行了

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[2005];
 
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_) {
        cin >> n;
        int x = 0, y = 0;
        rep (i, 1, (n << 1)) {
            cin >> a[i];
            if (i == (n << 1)) {
                if (x == 0 && y == 0) continue;
            } 
 
            if (i == (n << 1)) continue;
 
            if (a[i] & 1) {
                if (x == 0) x = i;
                else cout << x << ' ' << i << '\n', x = 0;
            } else {
                if (y == 0) y = i;
                else cout << y << ' ' << i << '\n', y = 0;
            }
        }
    }
    return 0;
}

C

sg函数不会写, 但和石子游戏差不多, 直接贪心也可

很明显 n = 1, A胜, n = 2 B胜

且 n & 1 == 1, 先手必胜

看看 n = 6, 6 -> 2 A胜

一个数可以因数分解成 a个奇数 和 b个偶数相乘(不考虑 1 这个因子)

通过找规律发现

1.n = 1 必败态

2.n = 2 必胜态

3.n & 1 == 1( a > 0, b = 0), 跳1, 必胜态 (n != 1)

4.当 a = 0, b > 0, A只能进行 -1 操作, 跳3, 必败态

5.当 a = 1, b = 1且这个数必为2, 要么跳2, 要么跳3, 必败态

6.当 a > 1, b = 1, 跳4, 必胜态

7.当 a > 1, b > 1, 跳4, 必胜态

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[2005];
 
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_) {
        cin >> n;
        if (n != 2 && n % 2 == 0 && (n >> 1) & 1) {
            bool flag = 0; m = n >> 1;
            for (int i = 3; i * i <= m; i += 2)
                if (m % i == 0) { flag = 1; break; }
            if (flag) cout << "Ashishgup\n";
            else cout << "FastestFinger\n";
        }
        else if (n == 1) cout << "FastestFinger\n";
        else if (n & 1 || n == 2) cout << "Ashishgup\n";
        else if ((n & n - 1) == 0) cout << "FastestFinger\n";
        else cout << "Ashishgup\n";
    }
    return 0;
}

D

看一眼 a[i] 和 n 的范围, 就应该向 二分想

二分, 按要求找check就行

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
int a[N];
bool b[N], c[N];
 
bool check() {
    int p = 0;
    rep (i, 1, n)
        if (p < k && c[p + 1] >= b[i]) ++p;
        else if (p == k) break;
 
    return p == k; 
}
 
bool solve(int mid) {
    rep (t, 0, 1) {
        rep (i, 1, n) c[i] = (i & 1) != t;
 
        if (check()) return true;
    }
    return false;
}
 
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    rep (i, 1, n) cin >> a[i];
 
    int l = 1, r =  1e9 + 1;
    while (l < r) {
        int mid = (r + l) >> 1;
        rep (i, 1, n) b[i] = a[i] > mid;
 
        if (solve(mid)) r = mid;
        else l = mid + 1; 
    }
 
    cout << l;
    return 0;
}

E

想象个容器, 放0, 放1, 按下标交叉放, 这个容器内的所有数一次旋转就可以弄好

查需要多少个这样的容器就行

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=a;i>=(b);--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e6 + 5;
 
int n, m, _, k;
char a[N], b[N];
 
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> a + 1 >> b + 1;
 
    int x = 0, y = 0, c = 0, d= 0;
    rep (i, 1, n) {
        if (a[i] == b[i]) continue;
 
        if (b[i] - '0') {
            ++x;
            d ? --d, ++c : ++c;
        } else {
            ++y;
            c ? --c, ++d : ++d;
        }
    }
 
    cout << (x == y ? c + d : -1);
 
    return 0;
}

F

找时间补吧

比赛时间写不出来的, 图论太差, 让我想起了, 上次交互提看题面1h时的 恐惧