洛谷 P1553 数字反转(升级版)
题目链接
思路
模拟题,采纳字符串输出,通过查找字符串中是否含有'/''.''%'字符,来分类解决。
代码
/* * @Description: * @Author: 多多 * @Date: 2020-10-26 10:01:01 * @LastEditTime: 2020-10-26 12:06:27 * @LastEditors: 多多 */#include <bits/stdc++.h>using namespace std;int main(){ freopen("./INPUT/P1553.in", "r", stdin); ios::sync_with_stdio(false); cin.tie(0); string s; cin >> s; if (s.find('.') != string::npos) { int index = s.find('.'); reverse(s.begin(), s.begin() + index); reverse(s.begin() + index + 1, s.end()); while (s[0] == '0' && s[1] != '.') { s.erase(0, 1); } while (s[s.length() - 1] == '0' && s[s.length() - 2] != '.') { s.erase(s.length() - 1, 1); } } else if (s.find('/') != string::npos) { int index = s.find('/'); reverse(s.begin(), s.begin() + index); reverse(s.begin() + index + 1, s.end()); while (s[0] == '0' && s[1] != '/') { s.erase(0, 1); } index = s.find('/'); while (s[index + 1] == '0') { s.erase(index + 1, 1); } } else if (s.find('%') != string::npos) { reverse(s.begin(), s.end() - 1); while (s[0] == '0' && s.length() != 2) { s.erase(0, 1); } } else { reverse(s.begin(), s.end()); while (s[0] == '0' && s.length() != 1) { s.erase(0, 1); } } cout << s << endl; return 0;}