Educational Codeforces Round 7

Educational Codeforces Round 7

https://codeforces.com/contest/622/problems
3/6: ABD

A. Infinite Sequence

水题

#include <bits/stdc++.h>

using namespace std;

int main () {
    long long x;
    cin >> x;
    long long n = sqrt (2 * x), m = (n + 1) * n / 2;
    //cout << m << ' ';
    if (x <= m) {
        cout << n - (m - x);
    }
    else {
        cout << x - m;
    }
}

B. The Time

水题

#include <bits/stdc++.h>

using namespace std;

int main () {
    int hh, mm, dx;
    scanf ("%d:%d", &hh, &mm);
    scanf ("%d", &dx);
    //cout << hh << ' ' << mm << ' ' << dx << endl;
    mm += dx;
    hh += mm / 60, mm %= 60, hh %= 24;
    //cout << hh << ' ' << mm;
    cout << setw(2) << setfill('0') << hh  << ':';
    cout << setw(2) << setfill('0') << mm << endl;
}

C. Not Equal on a Segment

这题思路很难想，个人想不到pos[]数组要这样做
一开始二分找，超时了

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5, M = 1e6 + 5;
int a[N], n, m;
int pos[N]; //i前面第一个相同的的下标

int main () {
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf ("%d", &a[i]);
        if (a[i] == a[i-1])     pos[i] = pos[i-1];
        else    pos[i] = i;
    }
    while (m --) {
        int l, r, x;
        scanf ("%d%d%d", &l, &r, &x);
        if (a[r] != x)  printf ("%d\n", r);
        else {
            if (pos[r] <= l) printf ("-1\n");
            else   printf ("%d\n", pos[r] - 1);
        }
    }
}

//找到[l,r]区间内不等于x的数的下标
//[1,1e6]

D. Optimal Number Permutation

非常开心！做出了一道1900
构造题，观察式子：

\(i,n\)都是固定的，所以就想一下怎么安排能够使得 \(d_i\) 的值近可能靠近 \(n-i\)。不妨设两个 \(i\) 之间的距离间隔即 \(d_i=n-i\)，则可以像下图一样构造：
（最后的 \(n\) 无论如何摆放贡献都是0，所以只需补最后一位的空即可）

#include <bits/stdc++.h>

using namespace std;
const int N = 5e5 + 5, M = N * 2;
int n, a[M];

int main () {
    cin >> n;
    for (int i = 1, j = 1; i <= n; i += 2, j++)    a[j] = a[n-j+1] = i;
    for (int i = 2, j = 1; i <= n; i += 2, j++)    a[n+j] = a[2*n-j] = i;
    a[2 * n] = n;
    for (int i = 1; i <= 2 * n; i++)    cout << a[i] << ' ';
}

/*
1: 1,   n
2: n+1, 2*n-1
3: 2,   n-1
4: n+2, 2n-2
...
*/

E. Ants in Leaves

因为蚂蚁是同时向上爬的，所以只需分治处理每颗子树，记录最大时间即可。递归处理每一个叶子的高度，然后排序求解。同一子树内，相同高度的要再等一下（因为不可以同时走到一个点上），可以 \(d[j] = max (d[j-1] + 1, d[j])\)这样更新，记录最大值。

#include <bits/stdc++.h>

using namespace std;
const int N = 5e5 + 5, M = N * 2;
int h[N], e[M], ne[M], idx;
int n, d[N], len, ans;

void add (int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs (int u, int fa, int dep) {
    bool isleaf = true;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        isleaf = false;
        dfs (j, u, dep + 1);
    }
    if (isleaf)     d[len++] = dep;
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        add (a, b), add (b, a);
    }
    for (int i = h[1]; ~i; i = ne[i]) {
        len = 0;
        dfs (e[i], 1, 1);
        sort (d, d + len);
        for (int j = 1; j < len; j++) {
            d[j] = max (d[j-1] + 1, d[j]);
        }
        ans = max (ans, d[len-1]);
    }
    cout << ans << endl;
}

F. The Sum of the k-th Powers

拉格朗日差值
原理：https://oi-wiki.org/math/poly/lagrange/ （公式推来推去）
题解：https://www.luogu.com.cn/blog/253608/solution-cf622f (和oiwiki上差不多)

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

using namespace std;
const int N = 1000005, p = 1000000007;
int n, k, ans, now;
int a[N], f[N], g[N];

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

void solve() {
    now = f[0] = g[0] = 1;
    for (int i = 1; i <= k + 2; i++)    now = now * (n - i) % p, f[i] = f[i - 1] * i % p, g[i] = -g[i - 1] * i % p;
    for (int i = 1; i <= k + 2; i++)    ans = (ans + a[i] * now % p * qmi(n - i, p - 2) % p * qmi(f[i - 1] * g[k + 2 - i] % p, p - 2)) % p;
}

signed main() {
    cin >> n >> k;
    for (int i = 1; i <= k + 2 and i <= n; i++)    a[i] = (a[i - 1] + qmi(i, k)) % p;
    if (n <= k + 2)    return cout << a[n], 0;
    solve();
    cout << (ans + p) % p;
}