(s1,0,16);//转成10进制 int b = stoi(s2,0,16); int c = stoi(s3,0,16); vector num =...ans; for(i = 0; i < 16; ++i) for(j = 0; j < 16; ++j) for(k = 0; k < 16; ++k) { x = stoi...(num[i],0,16);//转成10进制 y = stoi(num[j],0,16); z = stoi(num[k],0,16); sim = -(...(s1,0,16);//转成10进制 int b = stoi(s2,0,16); int c = stoi(s3,0,16); vector num =...= INT_MIN, sim1, sim2, sim3; string ans1, ans2, ans3; for(int i = 0, n; i < 16; ++i) { n = stoi
string strRight = strRequest.substr(nFind + 1, strRequest.size() - nFind); int nL = stoi...(strLeft); int nR = stoi(strRight); int nResult = nL + nR; return...(strLeft); int nR = stoi(strRight); int nResult = nL - nR; return...(strLeft); int nR = stoi(strRight); int nResult = nL * nR; return...(strLeft); int nR = stoi(strRight); if (0 == nR) {
= 0) || (x % 100 == 0 && x % 400 == 0)) return 1; else return 0; } int Stoi(string s){ //蓝桥杯不支持将...return sum; } bool is_t(string x, int p){ //判断该年份是否合法,即月份和天数是否合法 int sum = 0; int m = Stoi...(a.substr(0, 2)), m2 = Stoi(b.substr(0, 2)); if(m1 == m2){ int d1 = Stoi(a.substr(2, 2)),...d2 = Stoi(b.substr(2, 2)); return d1 > d2; } else return m1 > m2; } string to_y(int...(t); int Y = year; int m = Stoi(k.substr(0, 2)); //取出月份 int d = Stoi(k.substr(2, 2)); /
字符串转换为十六进制数: 使用 std::stoi 函数将字符串转换为十六进制数。...std::string hexString = "1A"; int decimal = std::stoi(hexString, nullptr, 16); // stoi的第三个参数16表示用16进制表示...常用十进制数的处理 字符串转换为十进制数: 使用 std::stoi 函数将字符串转换为十进制数。...std::string decimalString = "42"; int decimal = std::stoi(decimalString); 十进制数转换为字符串: 使用 std::to_string
timeToint(string& s, int g = 5, bool end = false) { // 例如 2017:01:01:23:59:59 long long Year = stoi...(s.substr(0,4)); long long Month = stoi(s.substr(5,2)); long long Day = stoi(s.substr...(8,2)); long long Hour = stoi(s.substr(11,2)); long long Minute = stoi(s.substr(14,2)...); long long Second = stoi(s.substr(17,2)); long long t; if(g==5)
} int daysBetweenDates(string date1, string date2) { int y1,y2,m1,m2,d1,d2; y1=stoi...(date1.substr(0,4)); y2=stoi(date2.substr(0,4)); m1=stoi(date1.substr(5,2));...m2=stoi(date2.substr(5,2)); d1=stoi(date1.substr(8,2)); d2=stoi(date2.substr(8,2));
Input: 3 167334 2333 12345678 Sample Output: Yes No No 解题思路: 输入string型的数字str,数字的位数为K(题目保证了K是偶数),然后通过stoi...string str; cin >> str; int K = str.length(); //数字Z的位数,题目保证了K是偶数 int Z = stoi...(str); //string型强制转换成int型 int A = stoi(str.substr(0,K/2)); //A是数字Z的前K/2位数字 int B...= stoi(str.substr(K/2)); //B是数字Z的后K/2位数字 if((A*B !
val; if(val == "N"){ return NULL; } TreeNode* root = new TreeNode(stoi...NULL; istringstream in(data); string s; in >> s; TreeNode* root = new TreeNode(stoi...N') node->left = NULL; else { node->left = new TreeNode(stoi...) node->right = NULL; else { node->right = new TreeNode(stoi
3; d++){ if(a + b + c + d == s.size()){ int A = stoi...(s.substr(0, a)); int B = stoi(s.substr(a, b));...int C = stoi(s.substr(a+b, c)); int D = stoi(s.substr(a+b+c, d));
stoi(version1) : stoi(version1.substr(0, index1)); int v21 = index2 == -1 ?...stoi(version2) : stoi(version2.substr(0, index2)); if(v11 > v21) return 1; else if(v11
for (int d = 1; d < 4; ++d) if (a + b + c + d == s.size()) { int A = stoi...(s.substr(0, a)); int B = stoi(s.substr(a, b)); int C = stoi(s.substr(a +...b, c)); int D = stoi(s.substr(a + b + c, d)); if (A <= 255 && B <= 255 &
string date) { int monthDays[12] = {31,28,31,30,31,30,31,31,30,31,30,31}; int year = stoi...(date.substr(0,4)), month = stoi(date.substr(5,2)), day = stoi(date.substr(
= countbits(low), num, delta; string s19 = "1234567890", s11 = "1111111111"; num = stoi....substr(0,bits)); vector ans; while(num <= high) { delta = stoi...ans.push_back(num); num += delta; } bits++; num = stoi
更新:把atoi(Pa.c_str())换成了stoi(Pa); 原谅我当时并不知道有stoi()这个函数。...(); i++) { if(B[i] == Db) { Pb += Db; } } int sum = stoi...(Pa) + stoi(Pb); //string型强制转换成int型再相加 cout << sum << endl; return 0; } 解法三思路&AC代码: 其实这三种解法思路都差不多
= string::npos) { int found = params_str.find(op); res.push_back(stoi(params_str.substr...(0, found))); params_str = params_str.substr(found + 1); } res.push_back(stoi(params_str...res; } int main() { // 客户预算 customer budget string mStr; getline(cin, mStr); int M = stoi
解题思路:通过"+"来对复数进行拆分,在C++中,atoi是将const char*类型转为int类型,而stoi是将string类型转为int类型。...stoi(num1) : ; int res2 = num2.length() != ?...stoi(num2) : ; return {res1, res2}; } }; 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems...vector& timePoints) { auto getMin = [](string& t1, string& t2) { int i1 = stoi...(t1.substr(, )) * + stoi(t1.substr(, )); int i2 = stoi(t2.substr(, )) * + stoi(t2.substr
y * x); else st.push(y / x); } else st.push(stoi...(tokens[i]));//stoi将string字符串变成十进制数字 } return st.top(); } };
1: 12345 2 输出样例 1: 3.66 输入样例 2: 12345 5 输出样例 2: 1.00 作者: 陈越 单位: 浙江大学 时间限制: 400 ms 内存限制: 64 MB 感谢柳婼教的stoi...; int a; cin>>a; string s1=s.substr(s.length()-a); s1=s1+s.substr(0,s.length()-a); int c,d; c=stoi...(s); d=stoi(s1); printf("%.2f",1.0*d/c); return 0; }
3 || tmp.size() == 0 || tmp.size() > 3 || (tmp.size() > 1 && tmp[0] == '0') || stoi...seg.push_back(tmp); } if(seg.size() < 4){ valid = false; } if (valid) { int f = stoi...else if (f >= 1 && f < 126) { cout << "A_address" << endl; }else if(f == 126 && stoi...(seg[1]) == 0 && stoi(seg[2]) == 0 && stoi(seg[3]) == 0){ cout << "A_address" << endl;
文本使用one-hot 编码步骤: 根据语料库创建 词典(vocabulary),并创建词和索引的 映射(stoi,itos); 将句子转换为用索引表示; 创建OneHot 编码器; 使用OneHot...reverse=True) # 词典 self.vocab = set(word_counter_sort) # 创建词和索引的映射 self.stoi...{} self.itos = {} for index, word in enumerate(word_counter_sort): self.stoi...self.vocab)) for index, item in enumerate(sentence.split()): result[index][self.stoi...# 单词编码 def word_encoder(self, word): result = [0] * len(self.vocab) result[self.stoi
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