PAT甲1038 Recover the smallest number
1038 Recover the Smallest Number (30 分)
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
题意:
给定n个数,要求任意组合使得组合得到的数最小。
思路:
之前做过一道类似的。这道题给的样例好啊。
表面上看好像是应该将输入的数字存成字符,按照字典序从小到大输出。但是观察样例会发现并不是这样的。
因为单个的字典序最小并不代表组合了之后的字典序最小。
应该要写一个cmp, 用的是组合之后字典序最小的排前面
还有要注意都是0的时候要记得最后输出一个0
#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f const int maxn = 1e4 + ;
string strnum[maxn];
int n; bool cmp(string a, string b)
{
return a + b < b + a;
} int main()
{
scanf("%d", &n);
for(int i = ; i < n; i++){
cin>>strnum[i];
}
//while(cin>>strnum[n++]);
sort(strnum, strnum + n, cmp);
bool flag = true;
for(int i = ; i < n; i++){
int l = strnum[i].length();
for(int j = ; j < l; j++){
if(flag && strnum[i][j] == '')continue;
if(flag && strnum[i][j] != ''){
flag = false;
}
printf("%c", strnum[i][j]);
}
}
if(flag){
printf("");
}
printf("\n");
return ;
}
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