Educational Codeforces Round 15

题目链接在这里

A. Maximum Increase

最长上升子串

#include <bits/stdc++.h>using namespace std;int n;int arr[100005];int main() {    cin >> n;    for (int i = 1; i <= n; i++)        cin >> arr[i];    int ans = 0;    int mx = 0;    for (int i = 1; i <= n; i++) {        if (arr[i] > arr[i - 1]) {            ans++;        } else {            mx = max(ans, mx);            ans = 1;        }    }    mx = max(mx, ans);    cout << mx << endl;    return 0;}

B. Powers of Two

n个数，问多少对 ai + aj = 2x(i<j)
map瞎搞。。

#include <bits/stdc++.h>using namespace std;int n;int arr[100005];map<int, int> vis;int main() {    cin >> n;    for (int i = 1; i <= n; i++) {        cin >> arr[i];        if (!vis[arr[i]]) vis[arr[i]] = 1;        else vis[arr[i]]++;    }    long long ans = 0;    for (int num = 2; num <= 1 << 30 && num > 0; num <<= 1) {        for (int i = 1; i <= n; i++) {            int a = arr[i];            if (!vis[num - a]) continue;            int b = num - a;            int t = vis[b];            if (a == b) ans += (long long)(t - 1);            else ans += t;        }    }    cout << ans / 2 << endl;    return 0;}

C. Cellular Network

直线上n个点，m座塔，每座塔都有相同的管辖范围r，问r最小是多少可以把n个点全部包含。

也就是说每个点左右距离r以内至少有一个灯塔，所以找到每个点距离最近的灯塔的距离，取最大值即可。

#include <bits/stdc++.h>using namespace std;int n, m;long long arr[100005];long long brr[100005];int main() {    cin >> n >> m;    for (int i = 1; i <= n; i++)        cin >> arr[i];    long long a;    for (int i = 1; i <= m; i++) {        cin >> brr[i];    }    brr[0] = -10000000000000000;    brr[m + 1] = 10000000000000000;    int idx = 0;    long long ans = 0;    for (int i = 1; i <= n; i++) {        while (idx < m && brr[idx + 1] < arr[i]) idx++;        ans = max(ans, min(arr[i] - brr[idx], brr[idx + 1] - arr[i]));    }    cout << ans << endl;    return 0;}

D. Road to Post Office

有d公里的距离要走，开车一公里a秒，但每开k公里要休息t秒，走路一公里b秒(a<b)，你一开始开车，并随时可以选择下来走路，问最短时间。

没啥好说的。。

#include <bits/stdc++.h>using namespace std;long long d, k, a, b, t;int main() {    cin >> d >> k >> a >> b >> t;    long long x = d / k;    long long y = d % k;    long long ans = 0;    if (!x) {        ans = d * a;    } else {        ans += k * a;        ans += min(t + y * a, y * b);        ans += min(a * k + t, b * k) * (x - 1);    }    cout << ans << endl;    return 0;}

E. Analysis of Pathes in Functional Graph

一个图，每个点都有且仅有一条出边，这条边有一个权值。
问从每个点开始走k步所经过的路径上的权值和与最小值。

用倍增的思想。
to[i][j]表示从i走2j步到达的点，
sum[i][j]表示从i走2j步路径上的权值和，
mi[i][j]表示从i走2j步路径上的权值最小值。
所以
to[i][j+1]=to[to[i][j]][j],
sum[i][j+1]=sum[i][j]+sum[to[i][j]][j],
mi[i][j+1]=min(mi[i][j],mi[to[i][j]][j]).

#include <bits/stdc++.h>using namespace std;const int MAXN = 100005;int n;long long k;int to[MAXN][55];long long val[MAXN][55];long long mi[MAXN][55];long long sum[MAXN], smi[MAXN];int idx[MAXN];int main() {    cin >> n >> k;    for (int i = 1; i <= n; i++) {        scanf("%d", &to[i][0]);        to[i][0]++;        idx[i] = i;    }    for (int i = 1; i <= n; i++) {        scanf("%lld", &val[i][0]);        mi[i][0] = val[i][0];    }    memset(smi, 0x3f, sizeof(smi));    int j = 0;    long long t = k;    while (k) {        for (int i = 1; i <= n; i++) {            val[i][j + 1] = val[i][j] + val[to[i][j]][j];            mi[i][j + 1] = min(mi[i][j], mi[to[i][j]][j]);            to[i][j + 1] = to[to[i][j]][j];        }        j++;        k >>= 1;    }    j--;    k = t;    while (k) {        while ((1ll << j) > k) j--;        for (int i = 1; i <= n; i++) {            sum[i] += val[idx[i]][j];            smi[i] = min(smi[i], mi[idx[i]][j]);            idx[i] = to[idx[i]][j];        }        k -= (1ll << j);    }    for (int i = 1; i <= n; i++) {        cout << sum[i] << ' ' << smi[i] << endl;    }    return 0;}