1073 Scientific Notation
part 1, 1.0
自己解法
#include <iostream>
using namespace std;
#include <string>int main()
{string s;cin >> s;string snum1 = "", snum2 = ""; // 表示前后两个数字int pos_split = s.find("E");int s_len = s.length();snum1 = s.substr(0, pos_split);snum2 = s.substr(pos_split + 1);int num2 = stoi(snum2);// 把sum1 继续拆分string sign = snum1.substr(0, 1);sign = (sign == "+") ? "" : "-";string b_dim = snum1.substr(1, 1);string a_dim = snum1.substr(3);string result = "";int shift_pos = 0; // 移动的位数if (num2 >= 0){int a_dim_len = a_dim.length();if (num2 >= a_dim_len){string s_zero;s_zero.assign(num2 - a_dim_len, '0');result += sign + b_dim + a_dim + s_zero;}elseresult += sign + b_dim + a_dim.substr(0, num2) + "." + a_dim.substr(num2);}else{string s_zero;s_zero.assign((-num2) - 1, '0');result += sign + "0." + s_zero + b_dim + a_dim;}cout << result << endl;system("pause");return 0;
}
大神解法
- 柳神
#include <iostream>
using namespace std;
int main() {string s;cin >> s;int i = 0;while (s[i] != 'E') i++;string t = s.substr(1, i-1);int n = stoi(s.substr(i+1));if (s[0] == '-') cout << "-";if (n < 0) {cout << "0.";for (int j = 0; j < abs(n) - 1; j++) cout << '0';for (int j = 0; j < t.length(); j++)if (t[j] != '.') cout << t[j];} else {cout << t[0];int cnt, j;for (j = 2, cnt = 0; j < t.length() && cnt < n; j++, cnt++) cout << t[j];if (j == t.length()) {for (int k = 0; k < n - cnt; k++) cout << '0';} else {cout << '.';for (int k = j; k < t.length(); k++) cout << t[k];}}return 0;
}
总结
- 由输入字符串不超过9999，则无法将其直接转为数字，而是在字符串之间进行转换
- 指数绝对值不超过9999，其可以转换为int
- stoi(string s): string 转 int
- 顺理成章的逻辑，不难