nodejs做网站容易被攻击吗深圳专门做seo的公司

很经典的大模拟题目 但是还不算难 大模拟题最需要注意的就是细节

写代码一定要考虑全面 并且要细心多debug 多打断点+STL库的熟练使用

istringstream真的处理字符串非常好用

注意解耦合思想 这样改代码debug更加清晰

https://www.acwing.com/problem/content/5724/

#include<bits/stdc++.h>
using namespace std;
#define THRESHOLD 1e-5  // 设置阈值为 1e-5
//给定方程式 思考是否可以配平0
//使用矩阵 元素个数*化合物个数
int n;
//得到化学反应方程式
vector<vector<int>> get_matrix(istringstream& iss) {map<int, map<string,int>> elements;//elements[数字][元素]=第几个化合物的元素的个数set<string> ee;//元素个数int m;//化合物个数iss >> m;//化合物个数for (int i = 0; i < m; i++) {string material;//每一个化合物iss >> material;string ss;//暂存元素for (int j = 0; j < material.size(); j++) {if (material[j] >= 'a' && material[j] <= 'z') {ss += material[j];}else {//是数字int num = material[j] - '0';while ((j + 1) < material.size() && material[j + 1] >= '0' && material[j + 1] <= '9') {num = num * 10 + (material[j + 1] - '0');j++;}elements[i][ss] += num;ee.insert(ss);ss = "";}}}//给元素排序vector<string> eee = vector<string>(ee.begin(), ee.end());//元素个数*化合物个数vector<vector<int>> matrix(eee.size(), vector<int>(m, 0));for (int i = 0; i < ee.size(); i++) {for (int j = 0; j < m; j++) {matrix[i][j] = elements[j][eee[i]];}}return matrix;
}vector<vector<double>> get_gauss_matrix(vector<vector<int>> matrix) {vector<vector<double>> double_matrix;for (const auto& row : matrix) {vector<double> new_row;for (int val : row) {new_row.push_back(static_cast<double>(val));}double_matrix.push_back(new_row);}int i = 0, j = 0;//(i,j)代表 当前矩阵左上角起始位置while (i < double_matrix.size() && j < double_matrix[0].size()) {int flag = 1;for (int k = i; k < double_matrix.size(); k++) {if (double_matrix[k][j] != 0) {flag = 0;break;}}//对子矩阵进行重复if (flag == 1) { j++; continue; }//否则else {if (double_matrix[i][j] == 0) {for (int k = i + 1; k < double_matrix.size(); k++) {if (double_matrix[k][j] != 0) {swap(double_matrix[i], double_matrix[k]);break;}}//pprint(double_matrix); cout << endl;}//遍历剩下所有行for (int k = i + 1; k < double_matrix.size(); k++) {double mul = double_matrix[k][j] / double_matrix[i][j];for (int z = j; z < double_matrix[0].size(); z++) {double_matrix[k][z] -= mul * double_matrix[i][z];}}}i++; j++;// 遍历矩阵并将小于阈值的元素设为0for (int i = 0; i < double_matrix.size(); ++i) {for (int j = 0; j < double_matrix[0].size(); ++j) {if (abs(double_matrix[i][j]) < THRESHOLD) {double_matrix[i][j] = 0;}}}}return double_matrix;
}
string judge(vector<vector<double>> gauss_matrix) {int num = gauss_matrix[0].size();//变量个数;int rank = 0;int flag = 1;for (int i = 0; i < gauss_matrix.size(); i++) {for (int j = 0; j < gauss_matrix[i].size(); j++) {//如果有非0的flag变成0if (gauss_matrix[i][j] != 0) {flag = 0;break;}        }if (flag == 1) {break;}else {rank++;flag = 1;}}if (rank < num) return "Y";return "N";
}
string solve(istringstream &iss) {vector<vector<int>> matrix=get_matrix(iss);//化学反应方程式vector<vector<double>> gauss_matrix = get_gauss_matrix(matrix);//获得高斯消元的行阶梯形式return judge(gauss_matrix);
}int main() {cin >> n;cin.ignore();  // 忽略换行符while (n--) {string s;getline(cin, s);istringstream iss(s);cout << solve(iss)<<endl;}return 0;
}