1038 Recover the Smallest Number (30 分)（贪心）

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