解题思路:
递归,应该第一次要个flag区别对待一下,我比较懒呢,就直接bypass了吧...
就是两个特殊数据,1和2要关照一下,虽然oj数据好像并没有测,有一点点水的
注意事项:
参考代码:
#define _CRT_SECURE_NO_WARNINGS #include <iostream> #include <string> #include <vector> #include <algorithm> #include <stdio.h> using namespace std; void solve(int curNum, string& str) { if (curNum == 0) { str += "2(0)"; return; } else if (curNum == 1) { str += "2"; return; } else if (curNum == 2) { str += "2(2)"; return; } else { str += "2("; vector<int> digitSum; int n = curNum, i = 0; while (n) { if (n % 2 != 0) digitSum.push_back(i); n /= 2; i += 1; } reverse(digitSum.begin(), digitSum.end()); for (vector<int>::iterator it = digitSum.begin(); it < digitSum.end(); it++) { solve(*it, str); if (it != --digitSum.end()) str += "+"; } str += ")"; } return; } string refine(int n) { if (n == 1) return "0"; else { string s; solve(n, s); return s.substr(2, s.length() - 3); } } int main(int argc, char** argv) { string s; int n = 0; cin >> n; cout << refine(n) << endl; return 0; }
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