当前位置:网站首页>竞赛题 2022-6-26
竞赛题 2022-6-26
2022-07-06 01:28:00 【Yake1965】
6104. 统计星号
class Solution:
def countAsterisks(self, s: str) -> int:
return "".join(s.split("|")[::2]).count("*")
flag, ans = 1, 0
for c in s:
if c == "*": ans += flag
elif c == "|": flag = 1 - flag
return ans
a = s.split('|')
return sum(a[i].count("*") for i in range(0, len(a), 2))
6107. 不同骰子序列的数目
第 i 次扔骰子的结果受第 (i - 1) 次和第 (i - 2) 次结果的限制。因此我们维护 f(i, a, b) 表示第 i 次扔骰子的结果为 a,第 (i - 1)次的结果为 b 的方案数。
为了转移,我们枚举第 (i - 2) 次的结果 c。根据题意,a,b,c 需要满足以下要求:
三个数互不相等;
gcd(a, b) = gcd(b, c) = 1。
只要同时满足这两个条件,f(i - 1, b, c) 就能转移给 f(i, a, b)。
第 300 场周赛
2325. 解密消息
class Solution {
public String decodeMessage(String key, String message) {
Map<Character, Character> map = new HashMap<>();
int i = 97;
for(char c : key.toCharArray()){
if(c == ' ') continue;
if(!map.containsKey(c)) map.put(c, (char)(i++));
}
map.put(' ', ' ');
StringBuilder sb = new StringBuilder();
for(char c : message.toCharArray()){
sb.append(map.get(c));
}
return sb.toString();
}
}
2326. 螺旋矩阵 IV
class Solution {
public int[][] spiralMatrix(int m, int n, ListNode head) {
int[][] ans = new int[m][n];
for(int[] row : ans) Arrays.fill(row, -1);
int up = 0, down = m - 1, left = 0, right = n - 1;
while(head != null){
for(int j = left; head != null && j <= right; j++){
ans[up][j] = head.val;
head = head.next;
}
up++;
for(int i = up; head != null && i <= down; i++){
ans[i][right] = head.val;
head = head.next;
}
right--;
for(int j = right; head != null && j >= left; j--){
ans[down][j] = head.val;
head = head.next;
}
down--;
for(int i = down; head != null && i >= up; i--){
ans[i][left] = head.val;
head = head.next;
}
left++;
}
return ans;
}
}
2327. 知道秘密的人数
f[i] 表示第 i 天新知道秘密的人数。
class Solution {
public int peopleAwareOfSecret(int n, int delay, int forget) {
int mod = 1000000007;
long s = 0;
long[] f = new long[n + 1];
f[1] = 1;
for(int i = 2; i <= n; i++){
if(i > delay) s += f[i-delay];
if(i > forget) s -= f[i-forget];
f[i] = s % mod;
}
long sum = 0;
for(int i = 0; i < forget; i++){
sum = (sum + f[n - i]) % mod;
}
return (int)sum;
}
}
2328. 网格图中递增路径的数目
边栏推荐
- leetcode刷题_验证回文字符串 Ⅱ
- Huawei Hrbrid interface and VLAN division based on IP
- WordPress collection plug-in automatically collects fake original free plug-ins
- Basic operations of databases and tables ----- default constraints
- Leetcode skimming questions_ Verify palindrome string II
- Mongodb problem set
- Yii console method call, Yii console scheduled task
- 【SSRF-01】服务器端请求伪造漏洞原理及利用实例
- Redis' cache penetration, cache breakdown, cache avalanche
- Condition and AQS principle
猜你喜欢
Opinions on softmax function
Vulhub vulnerability recurrence 74_ Wordpress
WordPress collection plug-in automatically collects fake original free plug-ins
ThreeDPoseTracker项目解析
基于DVWA的文件上传漏洞测试
【Flask】官方教程(Tutorial)-part2:蓝图-视图、模板、静态文件
晶振是如何起振的?
Test de vulnérabilité de téléchargement de fichiers basé sur dvwa
037 PHP login, registration, message, personal Center Design
Basic operations of database and table ----- set the fields of the table to be automatically added
随机推荐
【详细】快速实现对象映射的几种方式
ctf. Show PHP feature (89~110)
Mlsys 2020 | fedprox: Federation optimization of heterogeneous networks
Force buckle 9 palindromes
SPIR-V初窥
[Yu Yue education] Liaoning Vocational College of Architecture Web server application development reference
基於DVWA的文件上傳漏洞測試
internship:项目代码所涉及陌生注解及其作用
Format code_ What does formatting code mean
IP storage and query in MySQL
基于DVWA的文件上传漏洞测试
LeetCode 322. Change exchange (dynamic planning)
Nmap: network detection tool and security / port scanner
leetcode刷题_验证回文字符串 Ⅱ
【Flask】官方教程(Tutorial)-part1:项目布局、应用程序设置、定义和访问数据库
General operation method of spot Silver
【Flask】静态文件与模板渲染
Unity VR resource flash surface in scene
Pbootcms plug-in automatically collects fake original free plug-ins
Basic process and testing idea of interface automation