1038. Recover the Smallest Number (30)【排序】——PAT (Advanced Level) Practise

题目信息

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

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

解题思路

排序，注意排序的比较函数较奇特

AC代码

#include <cstdio>#include <string>#include <algorithm>using namespace std;string node[10005];bool cmp(const string s1, const string s2){    return s1 + s2 < s2 + s1;}int main(){    int n;    char ts[10];    scanf("%d", &n);    for (int i = 0; i < n; ++i){        scanf("%s", ts);        node[i] = ts;    }    sort(node, node + n, cmp);    string s;    for (int i = 0; i < n; ++i){        s += node[i];    }    n = 0;    while (n < s.size() && s[n] == '0') ++n;    if (n == s.size()){        printf("0\n");    }else{        printf("%s\n", s.substr(n).c_str());    }    return 0;}