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

#include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string str[10001], ans;
bool cmp(string a, string b){
        return a + b < b + a;
}
int main(){
    int N, index = -1;
    scanf("%d", &N);
    for(int i = 0; i < N; i++)
        cin >> str[i];
    sort(str, str + N, cmp);
    for(int i = 0; i < N; i++){
        ans += str[i];
    }
    int i = 0;
    while(ans[i] != '' && ans[i] == '0'){
        index = i;
        i++;
    }
    if(index != -1)
        ans.erase(0, index + 1);
    if(ans.length() == 0)
        cout << "0";
    else cout << ans;
    cin >> N;
    return 0;
}

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)