PAT (Advanced Level) Practise 1038 Recover the Smallest Number (30)

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

题意：给你n个数，将它们按一定顺序连起来，使得连起来后最小

解题思路：关键在于a+b<b+a，根据这个排序，然后还是注意下前导零

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;string s[10009];int n;bool cmp(string a, string b){return a + b < b + a;}int main(){while (~scanf("%d", &n)){for (int i = 0; i < n; i++) cin >> s[i];sort(s, s + n, cmp);int flag = 0;for (int i = 0; i < n; i++){for (int j = 0; s[i][j]; j++){if (!flag&&s[i][j] == '0') continue;flag = 1; printf("%c", s[i][j]);}}if (!flag) printf("0");printf("\n");}return 0;}