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
生词
| 英文 | 解释 |
|---|---|
| recover | 恢复 |
题目大意：
给一些字符串，求它们拼接起来构成最小数字的方式
分析：
贪心算法。让我们一起来见证cmp函数的强大之处！！不是按照字典序排列就可以的，必须保证两个字符串构成的数字是最小的才行，所以cmp函数写成return a + b < b + a;的形式，保证它排列按照能够组成的最小数字的形式排列。
因为字符串可能前面有0，这些要移除掉（用s.erase(s.begin())就可以了嗯string如此神奇~）。输出拼接后的字符串即可。
注意：
如果移出了0之后发现s.length() == 0了，说明这个数是0，那么要特别地输出这个0，否则会什么都不输出～
原文链接：https://blog.csdn.net/liuchuo/article/details/52264827
题解
#include <bits/stdc++.h>
using namespace std;
//注入灵魂的比较函数
bool cmp(string a,string b){
return a+b<b+a;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
int n;
cin>>n;
string str[n];
for(int i=0;i<n;i++){
cin>>str[i];
}
sort(str,str+n,cmp);
string s;
for(int i=0;i<n;i++){
s+=str[i];
}
while(s.length()!=0&&s[0]=='0'){
s.erase(s.begin());
}
if(s.length()==0) cout<<0;
else cout<<s;
return 0;
}
理论解释
