Source:

PAT A1038 Recover the Smallest Number (30 分)

Description:

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 (≤) followed by Nnumber 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

Keys:

  • 贪心

Code:

 /*
Data: 2019-07-23 18:47:04
Problem: PAT_A1038#Recover the Smallest Number
AC: 16:22 题目大意:
给几个数,求拼成的最小数
*/
#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
const int M=1e4+;
string s[M]; bool cmp(string a, string b)
{
return a+b < b+a;
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif int n;
scanf("%d", &n);
for(int i=; i<n; i++)
cin >> s[i];
sort(s,s+n,cmp);
string ans="";
for(int i=; i<n; i++)
ans += s[i];
while(ans.size()> && ans[]=='')
ans.erase(,);
cout << ans; return ;
}

PAT_A1038#Recover the Smallest Number的更多相关文章

  1. 把数组排成最小的数/1038. Recover the Smallest Number

    题目描述 输入一个正整数数组,把数组里所有数字拼接起来排成一个数,打印能拼接出的所有数字中最小的一个.例如输入数组{3,32,321},则打印出这三个数字能排成的最小数字为321323.   Give ...

  2. 1038. Recover the Smallest Number (30) - 字符串排序

    题目例如以下: Given a collection of number segments, you are supposed to recover the smallest number from ...

  3. A1038. Recover the Smallest Number

    Given a collection of number segments, you are supposed to recover the smallest number from them. Fo ...

  4. PAT甲1038 Recover the smallest number

    1038 Recover the Smallest Number (30 分) Given a collection of number segments, you are supposed to r ...

  5. 1038 Recover the Smallest Number (30 分)

    1038 Recover the Smallest Number (30 分) Given a collection of number segments, you are supposed to r ...

  6. 1038. Recover the Smallest Number (30)

    题目链接:http://www.patest.cn/contests/pat-a-practise/1038 题目: 1038. Recover the Smallest Number (30) 时间 ...

  7. PAT 1038 Recover the Smallest Number[dp][难]

    1038 Recover the Smallest Number (30 分) Given a collection of number segments, you are supposed to r ...

  8. PAT 甲级 1038 Recover the Smallest Number

    https://pintia.cn/problem-sets/994805342720868352/problems/994805449625288704 Given a collection of ...

  9. pat1038. Recover the Smallest Number (30)

    1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...

随机推荐

  1. 公司-浪潮:浪潮/inspur

    ylbtech-公司-浪潮:浪潮/inspur 浪潮集团有限公司,即浪潮集团,是中国本土综合实力强大的大型IT企业之一,中国领先的云计算.大数据服务商.浪潮集团旗下拥有浪潮信息.浪潮软件.浪潮国际.华 ...

  2. jackson反序列化报错Unrecognized field , not marked as ignorable

    使用Jackson提供的json注解. @JsonIgnore注解用来忽略某些字段,可以用在Field或者Getter方法上,用在Setter方法时,和Filed效果一样.这个注解只能用在POJO存在 ...

  3. jquery给表格绑值

    jquery给表格绑值 直接上代码了 <!DOCTYPE html> <html> <head> <meta charset="UTF-8" ...

  4. openstack部署安装

    OpenStack实战 准备环境 controller 10.0.0.11 compute1 10.0.0.31 常用服务端口 mariadb:3306 memcached:11211 消息队列:56 ...

  5. NMS python实现

    import numpy as np ''' 目标检测中常用到NMS,在faster R-CNN中,每一个bounding box都有一个打分,NMS实现逻辑是: 1,按打分最高到最低将BBox排序 ...

  6. 新浪sina邮箱客户端配置

    接收协议:IMAP 接收邮箱服务器地址:imap.sina.com 端口:993 加密方法:TLS 发送协议:SMTP 发送服务器:smtp.sina.com 端口:465 加密方法:TLS

  7. python+tushare获取股票和基金每日涨跌停价格

    接口:stk_limit 描述:获取全市场(包含A/B股和基金)每日涨跌停价格,包括涨停价格,跌停价格等,每个交易日8点40左右更新当日股票涨跌停价格. 限量:单次最多提取4800条记录,可循环调取, ...

  8. pip安装任何包都出现问题

    <!DOCTYPE html> { margin: 0; padding: 0; } body { background: url(images/body_bg.png) repeat-x ...

  9. Error(10028):Can't resolve multiple constant drivers for net “ ” at **.v

    两个进程里都有同一个条件判断的话,会产生并行信号冲突的问题. 同一个信号不允许在多个进程中赋值,否则则为多驱动. 进程的并行性决定了多进程不同能对同一个对象进行赋值.

  10. CG-CTF CRYPTO部分wp

    1,easybase64解密得flag 2,keyboard键盘码,在键盘上画画得flag:areuhack 3,异性相吸根据提示,写脚本 with open('密文.txt')as a: a=a.r ...