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
解题思路:
将数字两两之间进行一定的规律进行排序即可,规律:
S1 + S2 < S2 + S1,即根据贪心策略,两个数字拼起来的数字总是最小的
这里直接用cmp函数编写规律,用sort排序即可
代码如下:
#include <algorithm>
#include <iostream>
#include <string>
#include <string.h>
using namespace std;
const int MAXN = 10001;
int Squence[MAXN];
string Numbers[MAXN];
bool cmp(string a, string b) {
return a + b < b + a; //string数字间的比较,会自动化成int进行比较
}
int main() {
int N;
cin >> N;
for (int i = 0; i < N; ++i) {
cin >> Numbers[i];
}
sort(Numbers, Numbers + N, cmp);
string ans;
for (int i = 0; i < N; ++i) {
ans += Numbers[i]; //将Numbers里的内容拼接进去
}
//抹0操作
while (ans.size() != 0 && ans[0] == '0') {
ans.erase(ans.begin()); //抹去开头
}
if (ans.size() == 0) {
cout << 0;
}
else {
cout << ans;
}
system("PAUSE");
return 0;
}