1038. Recover the Smallest Number (30)

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 (<=10000) 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. Do not output leading zeros.

Sample Input:

5 32 321 3214 0229 87

Sample Output:

22932132143287

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给出若干个数，求他们能组成的最小的数。其实就是用排序。关键是比较函数。两个数对比对应的位，如果有不一样的位，则这个位较小的那个数优先级较高（比如5和32，32优先级比5高，因为组成的数325比532小），如果直到某个数到了结尾还没分出大小，则要延长该数来对比（为什么延长看看例子就能意会），比如32和321，将32延长成323，故321优先级较高（这两个数能组成32321和32132，32132比较小）。有了比较函数就能进行排序，用sort函数。最后输出结果，注意前面的0不能输出。这里还要注意可能最后的结果可能为0。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;bool cmp(string s1,string s2){int n=s1.size(),m=s2.size();if(n<=m){while(n<m){s1=s1+s1.substr(0,(m-n<=n)?m-n:n);n=s1.size();}}else{while(m<n){s2=s2+s2.substr(0,(n-m<=m)?n-m:m);m=s2.size();}}for(int i=0;i<n;i++){if(s1[i]!=s2[i]) return s1[i]<s2[i];}return true;}int main(){int n;cin>>n;vector<string>nums(n);for(int i=0;i<n;i++){cin>>nums[i];}sort(nums.begin(),nums.end(),cmp);bool flg=true;for(int i=0;i<nums.size();i++){for(int j=0;j<nums[i].size();j++){if(flg&&nums[i][j]=='0') continue;cout<<nums[i][j];flg=false;}}if(flg) cout<<0;}

9.10更新

发现以前的cmp函数写复杂了，这里更新比较简单的。

bool cmp(const string& s,const string& t){return s+t<t+s;}