PAT典型题目（8）贪心

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最新推荐文章于 2022-08-19 05:48:02 发布

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通过给定的一组数字片段，找到能组合成的最小数字。例如，给定{32, 321, 3214, 0229, 87}，最小数字可能是0229-321-3214-32-87。程序通过排序和拼接实现数字恢复，并移除前导零。" 121070176,11101954,MySQL数据库操作详解,"['数据库', 'SQL', 'MySQL', '数据查询']

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1038. Recover the Smallest Number (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
const int maxn=10010;
string str[maxn];
bool cmp(string a,string b){
return a+b<b+a;
}
int main(){
int n;
scanf("%d",&n);
for(int i=0;i<n;i++){
cin>>str[i];
}
sort(str,str+n,cmp);
string ans;
for(int i=0;i<n;i++){
ans+=str[i];
}
while(ans.size()!=0&&ans[0]=='0'){
ans.erase(ans.begin());
}
if(ans.size()==0) cout<<0;
else cout<<ans;
return 0;
}