1038 Recover the Smallest Number （30 分）PAT (Advanced Level) Practice

1038 Recover the Smallest Number （30 分）

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

 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～

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#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
bool cmp0(string a, string b) {
    return a + b < b + a;
}
string str[10010];
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        cin >> str[i];
    sort(str, str + n, cmp0);
    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;
    cout << s;
    return 0;
}

#include<iostream>
#include <vector>
#include <algorithm>
#include<string>
using namespace std;

int cmp(string a,string b)
{
    return a+b<b+a;//升序排	
}
string str[10010];
int main() 
{
    int n;
    scanf("%d",&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=s+str[i]; 
     
    int j=0;
	for(;j<s.length();j++)
	if(s[j]!='0')break;
	
	if(j==s.length())cout<<0;
	else
	{
	for(;j<s.length();j++)
	cout<<s[j];	
	}
	
	return 0; 
  
}