【笨方法學PAT】1038 Recover the Smallest Number (30 分)
阿新 • • 發佈:2018-12-20
一、題目
Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.
Input Specification:
Each input file contains one test case. Each case gives a positive integer N (≤104) followed by N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the smallest number in one line. Notice that the first digit must not be zero.
Sample Input:
5 32 321 3214 0229 87
Sample Output:
22932132143287
二、題目大意
一串數字,組成最小的數。
三、考點
貪心演算法
四、注意
1、如果不知道這個神奇的cmp,解決起來估計特別麻煩吧。
五、程式碼
#include<iostream> #include<string> #include<vector> #include<algorithm> using namespace std; bool cmp(string s1, string s2) { return s1 + s2 < s2 + s1; } int main() { //read int n; cin >> n; vector<string> v(n); for (int i = 0; i < n; ++i) cin >> v[i]; //sort sort(v.begin(), v.end(), cmp); //output string s; for (int i = 0; i < n; ++i) s+=v[i]; int i = 0; while (s[i] == '0' && i < s.length()) ++i; if (i == s.length()) cout << 0 << endl; else { for (; i < s.length(); ++i) cout << s[i]; cout << endl; } system("pause"); return 0; }