A1038. Recover the Smallest Number
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
1 #include<cstdio> 2 #include<iostream> 3 #include<string> 4 #include<algorithm> 5 using namespace std; 6 string str[10001], ans; 7 bool cmp(string a, string b){ 8 return a + b < b + a; 9 } 10 int main(){ 11 int N, index = -1; 12 scanf("%d", &N); 13 for(int i = 0; i < N; i++) 14 cin >> str[i]; 15 sort(str, str + N, cmp); 16 for(int i = 0; i < N; i++){ 17 ans += str[i]; 18 } 19 int i = 0; 20 while(ans[i] != '\0' && ans[i] == '0'){ 21 index = i; 22 i++; 23 } 24 if(index != -1) 25 ans.erase(0, index + 1); 26 if(ans.length() == 0) 27 cout << "0"; 28 else cout << ans; 29 cin >> N; 30 return 0; 31 }
总结：
1、题意：将若干字符串拼接起来，使得组成的数字最小。如果直接按照字典序由小到大排列再拼接是不对的。正确做法：比较串a和串b的位置关系时应考虑，如果a+b < b + a，则说明a前b后所组成的数字更小。
2、输出结果要去掉前导0，但不能仅仅对str[0]去0，因为有可能存在str[N] = {"0000", "0000", "0000"}的情况，应全部拼接起来后统一去除0，当最终得到的字串长度为0时，需要人为输出一个0。
3、string去除子串：str.erase(首位置， 长度)
