#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;getline(cin, s);int sz = s.size();int cnt = 0;for (int i = 0; i < sz; i++){if (isspace(s[i]))continue;elsecnt++;}cout << cnt << endl;return 0;
}

isspace判斷是否是空白字符，即空格和換行

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;getline(cin, s);int cnt = 0;for (auto x : s){if (isspace(x))continue;elsecnt++;}cout << cnt << endl;return 0;
}

方法二：按照單詞讀取

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;int cnt = 0;while (cin >> s){cnt += s.size();        }cout << cnt << endl;return 0;
}

有時候處理?個字符串的時候，也不?定要?次性讀取完整個字符串，如果字符串中有空格的話，其實可以當做多個單詞，?次讀取。
cin >> s 會返回?個流對象的引?，即 cin 本?。在C++中，流對象（如 cin ）可以被?作布爾值來檢查流的狀態。如果流的狀態良好（即沒有發?錯誤），流對象的布爾值為 true 。如果發?錯誤（如遇到輸?結束符或類型不匹配），布爾值為 false 。
在 while (cin >> s) 語句中，循環的條件部分檢查 cin 流的狀態。如果流成功讀取到?個值， cin >> s 返回的流對象cin 將被轉換為true ，循環將繼續。如果讀取失敗（例如遇到輸?結束符或?法讀取到?個值）， cin >> s 返回的流對象 cin 將被轉換為 false ，循環將停?。

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);int n;cin >> n;string p1, p2;while (n--){cin >> p1 >> p2;if (p1 == p2)cout << "Tie" << endl;else if (p1 == "Rock" && p2 == "Scissors")cout << "Player1" << endl;else if (p1 == "Scissors" && p2 == "Paper")cout << "Player1" << endl;else if (p1 == "Paper" && p2 == "Rock")cout << "Player1" << endl;elsecout << "Player2" << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;getline(cin, s);for (auto &x : s){if ((x >= 'b' && x <= 'z') || (x >= 'B' && x <= 'Z'))x--;else if (x == 'a')x = 'z';else if (x == 'A')x = 'Z';}cout << s << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);int q = 0;int m = 0;string s;string str;int a, b;cin >> q;cin >> s;while (q--){cin >> m;switch(m){case 1:cin >> str;s += str;cout << s << endl;break;case 2:cin >> a >> b;s = s.substr(a, b);cout << s << endl;break;case 3:cin >> a >> str;s.insert(a, str);cout << s << endl;break;case 4:cin >> str;size_t n = s.find(str);if (n == string::npos)cout << -1 << endl;elsecout << n << endl;break;}}return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;cin >> s;cout << s.size();while (cin >> s){size_t n = s.size();cout << ',' << n;}cout << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;bool flag = true;while (cin >> s){if (flag){cout << s.size();flag = false;}else{size_t n = s.size();cout << ',' << n;  }}cout << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;while (cin >> s){int left = 0;int right = s.size() - 1;while (left < right){char tmp = s[left];s[left] = s[right];s[right] = tmp;left++;right--;}cout << s << endl;}return 0;
}

其實在C++的STL中，包含?個算法叫 reverse ，可以完成字符串的逆序（反轉）。需要的頭?件是 <algorithm>

void reverse (BidirectionalIterator first, BidirectionalIterator last);  
//first: 指向要反轉范圍的第?個元素的迭代器（也可以是地址）  
//last: 指向要反轉范圍的最后?個元素的下?個位置的迭代器（也可以是地址）（翻轉時不包括此元素）。

reverse 會逆序范圍 [first, last) 內的元素

string s = "abcdef";  
reverse(s.begin(), s.end());

#include <iostream>  
#include <algorithm>  
using namespace std;
int main()  
{  //反轉字符串  string s("abcdef");  reverse(s.begin(), s.end());  cout << s << endl;  //反轉數組  int arr[] = { 1,2,3,4,5,6,7,8,9,10 };  int size = sizeof(arr) / sizeof(arr[0]);  //對數組中的元素進?反轉  reverse(arr, arr+size);  for (auto e : arr)  {  cout << e << " ";  }  cout << endl;  return 0;  
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;while (cin >> s){reverse(s.begin(), s.end());cout << s << endl;}return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;cin >> s;string s1;s1 = s;reverse(s.begin(), s.end());if (s1 == s)cout << "yes" << endl;elsecout << "no" << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;cin >> s;int left = 0;int right = s.size() - 1;while (left < right){if (s[left] != s[right]){cout << "no" << endl;break;}left++;right--;}if (left >= right)cout << "yes" << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int c[26] = {1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4};
int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);string s;getline(cin, s);int cnt = 0;for (auto x : s){if (x == ' ')cnt++;elsecnt += c[x - 'a'];}cout << cnt << endl;return 0;
}

#include <bits/stdc++.h>
using namespace std;int main()
{ios::sync_with_stdio(false);cin.tie(nullptr);int i = 0;cin >> i;string op;string last;while(i--){string ans;cin >> op;int n1, n2;int r = 0;if (op == "a" || op == "b" || op == "c"){cin >> n1 >> n2;ans += to_string(n1);if (op == "a"){r = n1 + n2;ans += "+";ans += to_string(n2);ans += "=";ans += to_string(r);}else if (op == "b"){r = n1 - n2;ans += "-";ans += to_string(n2);ans += "=";ans += to_string(r);}else{r = n1 * n2;ans += "*";ans += to_string(n2);ans += "=";ans += to_string(r);                }last = op;}else{n1 = stoi(op);ans += to_string(n1);cin >> n2;if (last == "a"){r = n1 + n2;ans += "+";ans += to_string(n2);ans += "=";ans += to_string(r);}else if (last == "b"){r = n1 - n2;ans += "-";ans += to_string(n2);ans += "=";ans += to_string(r);}else{r = n1 * n2;ans += "*";ans += to_string(n2);ans += "=";ans += to_string(r);                }}cout << ans << endl;cout << ans.size() << endl;}return 0;
}

#include <iostream>  
#include <string>  
using namespace std;  
int main()  
{  int n = 0;  cin >> n;  string op;  string num1;  string num2;  string last;  int ret = 0;  while (n--)  {  string ans;  cin >> op;  if (op == "a" || op == "b" || op == "c")  {  cin >> num1 >> num2;  int n1 = stoi(num1);  int n2 = stoi(num2);  ans += num1;  if (op == "a")  ret = n1 + n2, ans += "+";  else if (op == "b")  ret = n1 - n2, ans += "-";  else  ret = n1 * n2, ans += "*";  last = op;  }  else  {  num1 = op;  cin >> num2;  int n1 = stoi(num1);  int n2 = stoi(num2);ans += num1;  if (last == "a")  ret = n1 + n2, ans += "+";  else if (last == "b")  ret = n1 - n2, ans += "-";  else  ret = n1 * n2, ans += "*";  }  ans += (num2 + "=" + to_string(ret));  cout << ans << endl;  cout << ans.size() << endl;  }  return 0;  
}