begin wordpress/赣州seo外包

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前言

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整体评价

还是E稍微有点意思，新周赛好像比预期要简单一些, ^_.

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珂朵莉 牛客周赛专栏

珂朵莉 牛客小白月赛专栏

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A. 小红的新周赛

思路: 模拟

#include <bits/stdc++.h>using namespace std;int main() {int res = 0;for (int i = 0; i < 6; i++) {int v;cin >> v;res += v;}cout << res << endl;return 0;
}

B. 小红的字符串

思路: 计数+模拟

引入 26 * 26的状态，进行计数

这有个好处，就是天然排序，避免大内存存字符串并排序

#include <bits/stdc++.h>using namespace std;int main() {// 26 * 26天然保序int cnt[26][26] = {0};string s;cin >> s;int n = s.length();for (int i = 0; i < n - 1; i++) {int p1 = s[i] - 'a';int p2 = s[i + 1] - 'a';cnt[p1][p2]++;}for (int i = 0; i < 26; i++) {for (int j = 0; j < 26; j++) {string ts = "";ts.push_back((char)(i + 'a'));ts.push_back((char)(j + 'a'));for (int t = 0; t < cnt[i][j]; t++) {cout << ts << endl;}}}}

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C. 小红的炸砖块

思路: 模拟

引入保存每列高度的数组，然后模拟即可

#include <bits/stdc++.h>using namespace std;int main() {int n, m, k;cin >> n >> m >> k;vector<int> cols(m, n);for (int i = 0; i < k; i++) {int r, c;cin >> r >> c;if (cols[c - 1] >= n - r + 1) {cols[c - 1]--;}}for (int i = 0; i < n; i++) {string r;for (int j = 0; j < m; j++) {r.push_back(cols[j] >= n - i ? '*' : '.');}cout << r << endl;}return 0;
}

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D. 小红统计区间（easy）

思路: 滑窗

非常典的一道滑窗题，双指针维护即可

#include <bits/stdc++.h>using namespace std;using int64 = long long;int main() {int n;int64 k;cin >> n >> k;vector<int64> pre(n + 1, 0);vector<int> arr(n);for (int i = 0; i < n; i++) {cin >> arr[i];pre[i + 1] = pre[i] + arr[i];}int64 res = 0LL;int j = 0;for (int i = 0; i < n; i++) {while (j <= i && pre[i + 1] - pre[j] >= k) {j++;}res += j;}cout << res << endl;return 0;
}

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E. 小红的好数组

思路: 找规律 + 组合数学

case给的非常良心

可以分类讨论，大概有4种类似的序列

arr1 = [偶数，偶数，偶数，偶数，偶数，偶数， …]

arr2 = [奇数，奇数，偶数，奇数，奇数，偶数，…]

arr3 = [奇数，偶数，奇数，奇数，偶数，奇数，…]

arr4 = [偶数，奇数，奇数，偶数，奇数，奇数，…]

奇数/偶数的分布，呈现强烈的规律

最终为这4种情况的组合方案和

#include <bits/stdc++.h>using namespace std;using int64 = long long;const int64 mod = (long)1e9 + 7;int64 ksm(int64 b, int64 v) {int64 r = 1LL;while (v > 0) {if (v % 2 == 1) {r = r * b % mod;}v /= 2;b = b * b % mod;}return r;
}int main() {int n, k;cin >> n >> k;int k2 = k / 2, k1 = k - k2;int64 r1 = ksm(k2, n);int64 r2 = ksm(k2, n/3) * ksm(k1, n - n/3) % mod;int64 r3 = ksm(k2, (n+1)/3) * ksm(k1, n - (n+1)/3) % mod;int64 r4 = ksm(k2, (n+2)/3) * ksm(k1, n - (n+2)/3) % mod;int64 res = (r1 + r2 + r3 + r4) % mod;cout << res << endl;return 0;
}

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F. 小红统计区间（hard）

思路: 离散化+树状数组

也是一道非常典的题

因为存在负数，所以滑窗的基础已经被破坏了

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;public class Main {static class BIT {int n;int[] arr;public BIT(int n) {this.n = n;this.arr = new int[n + 1];}int query(int p) {int res = 0;while (p > 0) {res += arr[p];p -= p & -p;}return res;}void update(int p, int d) {while (p <= n) {arr[p] += d;p += p & -p;}}}public static void main(String[] args) {AReader sc = new AReader();int n = sc.nextInt();long k = sc.nextLong();long[] arr = new long[n];long[] pre = new long[n + 1];for (int i = 0; i < n; i++) {arr[i] = sc.nextLong();pre[i + 1] = pre[i] + arr[i];}// 进行离散化TreeSet<Long> ids = new TreeSet<>();for (long v: pre) {ids.add(v);}int ptr = 0;TreeMap<Long, Integer> hp = new TreeMap<>();for (long kv: ids) {hp.put(kv, ++ptr);}BIT bit = new BIT(ptr);bit.update(hp.get(0l), 1);long res = 0;for (int i = 0; i < n; i++) {long p = pre[i + 1];// p - x >= k// x <= p - kMap.Entry<Long, Integer> ent = hp.floorEntry(p - k);if (ent != null) {res += bit.query(ent.getValue());}bit.update(hp.get(p), 1);}System.out.println(res);}staticclass 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}}

#include <bits/stdc++.h>using namespace std;
using int64 = long long;class BIT {
private:int n;vector<int> arr;
public:BIT(int n): n(n), arr(n + 1, 0) {}int query(int p) {int r = 0;while (p > 0) {r += arr[p];p -= p & -p;}return r;}void update(int p, int d) {while (p <= n) {arr[p] += d;p += p & -p;}}
};int main() {int n;int64 k;cin >> n >> k;vector<int64> pre(n + 1, 0LL);for (int i = 0; i < n; i++) {int v;cin >> v;pre[i + 1] = pre[i] + v;}set<int64> ts;for (int64 v: pre) {ts.insert(v);}int ptr = 0;map<int64, int> idMap;for (int64 v: ts) {idMap[v] = ++ptr;}int64 res = 0;BIT bit(ptr);bit.update(idMap[0], 1);for (int i = 0; i < n; i++) {int64 p = pre[i + 1];if (idMap.find(p - k) != idMap.end()) {res += bit.query(idMap[p - k]);} else {auto iter = idMap.lower_bound(p - k);if (iter != idMap.end()) {res += bit.query(iter->second - 1);} else {res += bit.query(ptr);}}bit.update(idMap[p], 1);}cout << res << endl;return 0;
}

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