AtCoder Beginner Contest 127

AtCoder Beginner Contest 127

https://atcoder.jp/contests/abc127/tasks
4/6: ABCD

A - Ferris Wheel

水题

#include <bits/stdc++.h>

using namespace std;

int main () {
    int a, b;
    cin >> a >> b;
    if (a <= 5) cout << 0;
    else if (a <= 12)   cout << b / 2;
    else    cout << b;
}

B - Algae

水题

#include <bits/stdc++.h>

using namespace std;

int main () {
    int r, d, ans;
    cin >> r >> d >> ans;
    for (int i = 1; i <= 10; i++) {
        ans = r * ans - d;
        cout << ans << endl;
    }
}

C - Prison

差分

#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 5;
int a[N], n, m, ans;

int main () {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int l, r;
        cin >> l >> r;
        a[l] ++, a[r+1] --;
    }
    for (int i = 1; i <= n; i++)    a[i] += a[i-1];
    for (int i = 1; i <= n; i++) {
        //cout << a[i] << ' ';
        if (a[i] == m)  ans ++;
    }
    cout << ans;
}

D - Integer Cards

二分
关键在于最终局面是固定的，所以每次贪心的把小的变为最大的。排序然后二分，二分边界不断右移是因为p数组降序排列之后保证了后面查找的都只会更小。

#include <bits/stdc++.h>
#define int long long

using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 5;
int a[N], n, m, ans;
pii p[N];

signed main () {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int i = 1; i <= m; i++)    cin >> p[i].second >> p[i].first;
    sort (p + 1, p + m + 1, greater<pii>());
    sort (a + 1, a + n + 1);
    int st = 1;
    //for (int i = 1; i <= n; i++)    cout << a[i] << ' ';cout << endl;
    for (int i = 1; i <= m; i++) {
        int val = p[i].first, cnt = p[i].second;
        int pos = lower_bound (a + st, a + n + 1, val) - a;
        int len = pos - st; //[st, pos-1]变为val
        //cout << st << ' ' << pos << endl; 
        ans += min (len, cnt) * val;
        st += min (len, cnt);
        //cout << st << ' ' << pos << endl; 
    }
    ans += a[st];
    while (st <= n)     ans += a[++st];
    cout << ans << endl;
}

//每次把最小的小于ci的bi个数字变为ci
//终局是固定的

E - Cell Distance

推式子，组合数学。
先选两个点，记横坐标之差为 $d_x=|x_1-x_2|$，则在长度为 $n$ 的行内可以选出差值为 $d_x$ 的点对有 $n-d_x$ 对，两点分别可能在 $m$ 列，所以这两点的行贡献为 $d_x\times(n-d_x)\times m^2$。同理，列贡献为 $d_y\times(m-d_y)\times n^2$。

还要在剩下的 $n\times m-2$ 个点中选 $k-2$ 个点，则再乘上 $C_{n\times m-2}^{k-2}$，则最终式子为：

$$(\sum_{i=1}^{n-1}i\times(n-i)\times m^2 + \sum_{i=1}^{m-1}i\times(m-i)\times n^2)\times C_{n\times m-2}^{k-2}$$

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 200005, mod = 1000000007;
int fact[N], infact[N], ans;

int qmi(int a, int k, int p) {
	int ans = 1;
	while (k) {
		if (k & 1)	ans = ans * a % p;
		k >>= 1;
		a = a * a % p;
	}
	return ans;
}

void init() {
    fact[0] = infact[0] = 1;
    for (int i = 1; i <= N - 3; i++) {
        fact[i] = fact[i - 1] * i % mod;
        infact[i] = infact[i - 1] * qmi (i, mod - 2, mod) % mod;
    }
}

int C(int a, int b) {
    return fact[a] * infact[a - b] % mod * infact[b] % mod;
}

signed main () {
    init ();
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i < n; i++)     ans += (i * (n - i) * m * m) % mod;
    for (int i = 1; i < m; i++)     ans += (i * (m - i) * n * n) % mod;
    ans = (ans % mod * C (n * m - 2, k - 2)) % mod;
    cout << ans << endl;
}

F - Absolute Minima

对顶堆。
利用绝对值的性质：
$|x-a|+|x-b|$ 就是 $ab$ 的长度，$x$ 在 $a$ 的左边或者 $b$ 的右边，只会让答案变大。所以有 $cnt$ 个 $a$ 时，若 $cnt$ 为偶数，则当 $x$ 取为 $a_{\frac{cnt}2}$​ 时答案最小。反之则为 $a_{\frac{cnt}2+1}$时最小.
则转化为经典题：动态维护第k小

#include <bits/stdc++.h>
#define int long long

using namespace std;
priority_queue <int> big; //大根堆堆顶即中位数
priority_queue <int, vector<int>, greater<int>> small;

signed main () {
    int q, sum = 0;
    cin >> q;
    while (q --) {
        int op;
        cin >> op;
        if (op == 2)    cout << big.top () << ' ' << sum << endl;
        else {
            int a, b; cin >> a >> b;
            sum += b, small.push (a), big.push (a);
            int ss = small.top (), bb = big.top ();
            small.pop (), big.pop ();
            if (ss < bb)      sum += bb - ss, swap (ss, bb);
            small.push (ss), big.push (bb);
        }
    }
}