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小金狮游戏ui培训_android studio安装教程_长沙seo_口碑营销例子

时间:2025/7/12 6:47:41来源:https://blog.csdn.net/l23456789mmmmm/article/details/144065306 浏览次数: 0次
小金狮游戏ui培训_android studio安装教程_长沙seo_口碑营销例子

1.3206. 交替组 I - 力扣(LeetCode)

class Solution {
public:int numberOfAlternatingGroups(vector<int>& colors) {int n = colors.size();int ans = 0;for (int i = 0; i < n; i++) {if (colors[i] != colors[(i - 1 + n) % n] &&colors[i] != colors[(i + 1 ) % n])ans++;}return ans;}
};

2.小F的永久代币卡回本计划 - MarsCode

v

#include <iostream>
using namespace std;int solution(int a, int b) {// write code hereint tmp = 0;if (a % b != 0)tmp = 1;return a / b + tmp;
}int main() {cout << (solution(10, 1) == 10) << endl;cout << (solution(10, 2) == 5) << endl;cout << (solution(10, 3) == 4) << endl;return 0;
}

3.数组元素之和最小化 - MarsCode

#include <iostream>
#include <vector>
#include <string>using namespace std;int solution(int n, int k) {int ans = 0,begin = k;while(n--){ans+=begin;begin+=k;}return ans;
}int main() {std::cout << (solution(3, 1) == 6) << std::endl;std::cout << (solution(2, 2) == 6) << std::endl;std::cout << (solution(4, 3) == 30) << std::endl;
}

4.最大矩形面积问题 - MarsCode

#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int solution(int n, std::vector<int> A) {// Edit your code herestack<int> s;int ans = 0;for (int i = 0; i < n; i++) {while (!s.empty() && A[i] < A[s.top()]) {int a = A[s.top()];s.pop();int b = s.empty() ? i : i - s.top() - 1;ans = max(ans, a * b);}ans = max(ans, A[i]);s.push(i);}while (!s.empty()) {int a = A[s.top()];s.pop();int b = s.empty() ? n : n - s.top() - 1;ans = max(ans, a * b);}// cout << ans << endl;return ans;
}
// [7,8,19,10,6,22,10,5,2,14]
// 15,5,22,19,5,20,13,12,12
int main() {// Add your test cases herestd::vector<int> A_case1 = std::vector<int>{1, 2, 3, 4, 5};std::cout << (solution(5, A_case1) == 9) << std::endl;std::vector<int> B_case1 = std::vector<int>{5, 4, 3, 2, 1, 6};std::cout << (solution(6, B_case1)) << std::endl;std::vector<int> C_case1 = std::vector<int>{4, 4, 4, 4};std::cout << (solution(4, C_case1)) << std::endl;std::vector<int> D_case1 = std::vector<int>{15, 5, 22, 19, 5, 20, 13, 12, 12};std::cout << (solution(9, D_case1)) << std::endl;return 0;
}

5.比赛配对问题 - MarsCode

#include <iostream>
#include <vector>
#include <string>using namespace std;int solution(int n) {// write code hereint ans = 0;while(n != 1){ans += n/2;n = n -n/2;}return ans;
}int main() {cout << (solution(7) == 6) << endl;cout << (solution(14) == 13) << endl;cout << (solution(1) == 0) << endl;return 0;
}

6.DNA序列编辑距离 - MarsCode

#include <iostream>
#include <string>
#include <vector>
using namespace std;
int solution(std::string dna1, std::string dna2) {// Please write your code hereint n1 = dna1.size(), n2 = dna2.size();vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 1000000));for (int i = 0; i <= n1; i++)dp[i][0] = i;for (int i = 0; i <= n2; i++)dp[0][i] = i;for (int i = 0; i < n1; i++) {for (int j = 0; j < n2; j++) {dp[i + 1][j + 1] = min(dp[i][j] + (dna1[i] != dna2[j]),min(dp[i + 1][j] + 1, dp[i][j + 1] + 1));// if(dna1[i] != dna2[j])//     dp[i + 1][j + 1]++;// cout << dp[i + 1][j + 1] << ' ';}// cout << endl;}return dp[n1][n2];
}int main() {//  You can add more test cases herestd::cout << (solution("AGT", "AGCT")) << std::endl;// std::cout << (solution("AACCGGTT", "AACCTTGG")) << std::endl;// std::cout << (solution("ACGT", "TGC")) << std::endl;std::cout << (solution("CCCCACTGGAAGTCCTCCTAT", "T")) << std::endl; // 20return 0;
}

7组成字符串ku的最大次数 - MarsCode

#include <iostream>
#include <vector>
#include <string>using namespace std;int solution(const string& s) {int k = 0,u = 0;for(auto e : s){if(e == 'k'|| e == 'K')k++;else if(e == 'u'|| e == 'U')u++;}return min(k,u);
}int main() {cout << (solution("AUBTMKAxfuu") == 1) << endl;cout << (solution("KKuuUuUuKKKKkkkkKK") == 6) << endl;cout << (solution("abcdefgh") == 0) << endl;
}

8.游戏排名第三大的分数 - MarsCode

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
bool fun(int a,int b){return a > b;
}
int solution(int n, std::vector<int> nums) {sort(nums.begin(),nums.end(),fun);int pos = 0;// cout << nums[0] <<endl;for(int i = 1;i < n;i++){if(nums[i] != nums[i - 1])pos++;if(pos == 2)return nums[i];}return nums[0];
}int main() {std::cout << (solution(3, {3, 2, 1})) << std::endl;std::cout << (solution(2, {1, 2})) << std::endl;std::cout << (solution(4, {2, 2, 3, 1})) << std::endl;return 0;
}

9完美偶数计数 - MarsCode

#include <iostream>
#include <vector>
#include <string>using namespace std;int solution(int n, int l, int r, vector<int>& a) {int ans = 0;for(auto e : a){if(e%2 == 0 && e>= l &&e <= r)ans++;}return ans;
}int main() {vector<int> a1 = {1, 2, 6, 8, 7};cout << (solution(5, 3, 8, a1) == 2) << endl;vector<int> a2 = {12, 15, 18, 9};cout << (solution(4, 10, 20, a2) == 2) << endl;vector<int> a3 = {2, 4, 6};cout << (solution(3, 1, 10, a3) == 3) << endl;
}

10RGB色值转换为整数值 - MarsCode

#include <iostream>
#include <string>
using namespace std;
int solution(std::string rgb) {// Please write your code hereauto pos0 = rgb.find('(');auto pos1 = rgb.find(',',pos0);auto pos2 = rgb.find(',',pos1 + 1);auto pos3 = rgb.find(')',pos2);int a = stoi(rgb.substr(pos0 + 1,pos1 - 1));int b = stoi(rgb.substr(pos1 + 2,pos2 - 1));int c = stoi(rgb.substr(pos2 + 2,pos3 - 1));// cout <<a <<' '<<b <<' '<<c<<endl;return a*256*256+b*256+c;
}int main() {//  You can add more test cases herestd::cout << (solution("rgb(192, 192, 192)") == 12632256) << std::endl;std::cout << (solution("rgb(100, 0, 252)") == 6553852) << std::endl;std::cout << (solution("rgb(33, 44, 55)") == 2174007) << std::endl;return 0;
}

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