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;
}