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【笨方法学PAT】1038 Recover the Smallest Number (30 分)
阅读量:133 次
发布时间:2019-02-26

本文共 1419 字,大约阅读时间需要 4 分钟。

一、题目

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 (≤10​4​​) 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、题目大意

一串数字,组成最小的数。

三、考点

贪心算法

四、注意

1、如果不知道这个神奇的cmp,解决起来估计特别麻烦吧。

五、代码

#include
#include
#include
#include
using namespace std;bool cmp(string s1, string s2) { return s1 + s2 < s2 + s1;}int main() { //read int n; cin >> n; vector
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;}

 

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