前言
整体评价
T5, T6有点意思，这场小白入门场，好像没真正意义上的签到，整体感觉是这样。
A. 召唤神坤
思路: 前后缀拆解
c++
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
// 请在此输入您的代码
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
vector<int> pre(n);
vector<int> suf(n);
pre[0] = -1;
for (int i = 1; i < n; i++) {
pre[i] = max(arr[i - 1], pre[i - 1]);
}
suf[n - 1] = -1;
for (int i = n - 2; i >= 0; i--) {
suf[i] = max(arr[i + 1], suf[i + 1]);
}
int res = 0;
for (int i = 1; i < n - 1; i++) {
res = max(res, (pre[i] + suf[i]) / arr[i]);
}
cout << res << endl;
return 0;
}B. 聪明的交换策略
思路: 模拟+枚举
c++
#include <bits/stdc++.h>
using namespace std;
int main()
{
// 请在此输入您的代码
int n;
cin >> n;
string s;
cin >> s;
// long long res = 1LL << 60;
long long acc1 = 0, acc2 = 0;
int t0 = 0, t1 = 0;
for (int i = 0; i < n;i++) {
char c = s[i];
if (c == '0') {
acc1 += (i - t0);
t0++;
} else {
acc2 += (i - t1);
t1++;
}
}
cout << min(acc1, acc2) << endl;
return 0;
}C. 怪兽突击
思路: 枚举
c++
#include <bits/stdc++.h>
using namespace std;
int main()
{
// 请在此输入您的代码
int n, k;
cin >> n >> k;
vector<int> arr(n), brr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = 0; i < n; i++) {
cin >> brr[i];
}
long long res = 1LL << 60;
long long pre = 0;
long long mv = arr[0] + brr[0];
for (int i = 0; i < n; i++) {
if (i + 1 > k) break;
pre += arr[i];
mv = min(mv, (long long)arr[i] + brr[i]);
res = min(res, pre + (k - i - 1) * mv);
}
cout << res << endl;
return 0;
}D. 蓝桥快打
思路: 二分
c++
#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
int main()
{
// 请在此输入您的代码
int t;
cin >> t;
while (t-- > 0) {
int a, b, c;
cin >> a >> b >> c;
// 可以二分的
int l = 1, r = b;
while (l <= r) {
int m = l + (r - l) / 2;
// *)
int times = (b + m - 1) / m;
if ((int64)(times - 1) * c < a) {
r = m - 1;
} else {
l = m + 1;
}
}
cout << l << endl;
}
return 0;
}E. 奇怪的段
思路: 单调队列优化的DP
这题只需要维护最大值就行，不需要维护单调队列
其核心是如下的公式
d p [ i ] [ j ] = max t = 0 t = j − 1 d p [ i − 1 ] [ t ] + ( p r e [ j + 1 ] − p r e [ t ] ) ∗ w [ i ] dp[i][j] = \max_{t=0}^{t=j-1} dp[i - 1][t] + (pre[j + 1] - pre[t]) * w[i] dp[i][j]=t=0maxt=j−1dp[i−1][t]+(pre[j+1]−pre[t])∗w[i]
公式拆解后
d p [ i ] [ j ] = max t = 0 t = j − 1 ( d p [ i − 1 ] [ t ] − p r e [ t ] ) ∗ w [ i ] ) + p r e [ j ] ∗ w [ i ] dp[i][j] = \max_{t=0}^{t=j-1}(dp[i - 1][t] - pre[t]) * w[i]) + pre[j] * w[i] dp[i][j]=t=0maxt=j−1(dp[i−1][t]−pre[t])∗w[i])+pre[j]∗w[i]
这样这个递推的时间代价为 O ( 1 ) O(1) O(1)，而不是 O ( n ) O(n) O(n)
这样总的时间复杂度为 O ( n ∗ k ) O(n * k) O(n∗k)
java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
AReader sc = new AReader();
int n = sc.nextInt(), p = sc.nextInt();
int[] arr = new int[n];
long[] pre = new long[n + 1];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
pre[i + 1] = pre[i] + arr[i];
}
int[] ws = new int[p];
for (int i = 0; i < p; i++) {
ws[i] = sc.nextInt();
}
// *)
long inf = Long.MIN_VALUE / 10;
long[][] dp = new long[p + 1][n];
for (int i = 0; i <= p; i++) {
Arrays.fill(dp[i], inf);
}
for (int i = 0; i < n; i++) {
dp[1][i] = pre[i + 1] * ws[0];
}
// O(n)
// dp[i - 1][j] - p * pre[j + 1] + p * pre[j]
for (int i = 2; i <= p; i++) {
long tmp = dp[i - 1][0] - ws[i - 1] * pre[1];
for (int j = 1; j < n; j++) {
dp[i][j] = tmp + ws[i - 1] * pre[j + 1];
tmp = Math.max(tmp, dp[i - 1][j] - ws[i - 1] * pre[j + 1]);
}
}
System.out.println(dp[p][n - 1]);
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
// public BigInteger nextBigInt() {
// return new BigInteger(next());
// }
// 若需要nextDouble等方法，请自行调用Double.parseDouble包装
}
}F. 小蓝的反击
思路: 滑窗 + 三指针
java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;
public class Main {
static List<int[]> split(int v) {
List<int[]> res = new ArrayList<>();
for (int i = 2; i <= v / i; i++) {
if (v % i == 0) {
int cnt = 0;
while (v % i == 0) {
v /= i;
cnt++;
}
res.add(new int[] {i, cnt});
}
}
if (v > 1) {
res.add(new int[] {v, 1});
}
return res;
}
// *)
static boolean check(int[][] pre, int s, int e, List<int[]> xx) {
for (int i = 0; i < xx.size(); i++) {
int tn = xx.get(i)[1];
if (pre[i][e + 1] - pre[i][s] < tn) return false;
}
return true;
}
public static void main(String[] args) {
AReader sc = new AReader();
int n = sc.nextInt();
int a = sc.nextInt();
int b = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
if (b == 1) {
System.out.println(0);
return;
}
List<int[]> facs1 = split(a);
int m1 = facs1.size();
List<int[]> facs2 = split(b);
int m2 = facs2.size();
// 三指针做法
// int[][] brr1 = new int[m1][n];
int[][] pre1 = new int[m1][n + 1];
// int[][] brr2 = new int[m2][n];
int[][] pre2 = new int[m2][n + 1];
for (int i = 0; i < n; i++) {
int v = arr[i];
for (int j = 0; j < m2; j++) {
int p = facs2.get(j)[0];
int tmp = 0;
while (v % p == 0) {
v /= p;
tmp++;
}
// brr2[j][i] = tmp;
pre2[j][i + 1] = pre2[j][i] + tmp;
}
v = arr[i];
for (int j = 0; j < m1; j++) {
int p = facs1.get(j)[0];
int tmp = 0;
while (v % p == 0) {
v /= p;
tmp++;
}
// brr1[j][i] = tmp;
pre1[j][i + 1] = pre1[j][i] + tmp;
}
}
long res = 0;
int k1 = 0, k2 = 0;
for (int k3 = 0; k3 < n; k3++) {
// 找到不满足的点为止
while (k1 <= k3 && check(pre1, k1, k3, facs1)) {
k1++;
}
//
while (k2 <= k3 && check(pre2, k2, k3, facs2)) {
k2++;
}
// res += Math.min(k1, k2);
// 0 - k1 - 1
// k2 -> n
res += (k2 <= k1 - 1) ? (k1 - k2): 0;
}
System.out.println(res);
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
// public BigInteger nextBigInt() {
// return new BigInteger(next());
// }
// 若需要nextDouble等方法，请自行调用Double.parseDouble包装
}
}写在最后
