(PAT 1038) Recover the Smallest Number(贪心算法)

Myth丶恋晨 2022-04-10 14:09 201阅读 0赞

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;
    }