PAT A1038 Recover the Smallest Number

原题

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 (≤104) 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

代码长度限制

16 KB

时间限制

200 ms

内存限制

64 MB

栈限制

8192 KB

解法

使用将string的结果合并后转为long long类型，比较合并后哪个字符串在前组成的数字更小 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;
const int N = 1e4 + 10;
int res[N];
vector<string> num;

bool cmp(string s1, string s2) {
    long long i1 = strtoll((s1 + s2).c_str(),  NULL, 10);
    long long i2 = strtoll((s2 + s1).c_str(),  NULL, 10);
    return i1 < i2;
}
int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i ++) {
        string s;
        cin >> s;
        num.push_back(s);
    }


    sort(num.begin(), num.end(), cmp);
    bool flag = false;
    for (int i = 0; i < n; i ++) {
        if (!flag && num[i][0] == '0') {
            int k = 0;
            while (k < num[i].size() && num[i][k] == '0') k ++;
            if (k < num[i].size()) { // 不等于0
                flag = true;
                while (k < num[i].size()) {
                    cout << num[i][k ++];
                }
            }
            // 等于0，不处理
        }
        else {
            flag = true;
            cout << num[i];
        }
    }
    if (!flag) { // 全为0
        cout << 0;
    }
    return 0;
}