• 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
    CHEN, Yue

    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
    

    提交代码

    学习网址:http://blog.csdn.net/sinat_29278271/article/details/48047877

    里面的传递性是可以证明的

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<queue>
     6 #include<vector>
     7 #include<cmath>
     8 #include<string>
     9 using namespace std;
    10 vector<string> v;
    11 bool cmp(string a,string b){
    12     return a+b<b+a;
    13 }
    14 int main(){
    15     //freopen("D:\INPUT.txt","r",stdin);
    16     int n,i;
    17     scanf("%d",&n);
    18     string s;
    19     for(i=0;i<n;i++){
    20         cin>>s;
    21         v.push_back(s);
    22     }
    23     sort(v.begin(),v.end(),cmp);
    24     s="";
    25     for(i=0;i<n;i++){
    26         s=s+v[i];
    27     }
    28     for(i=0;i<s.length();i++){
    29         if(s[i]!='0'){
    30             break;
    31         }
    32     }
    33 
    34     if(i==s.length()){
    35         printf("0
    ");
    36     }
    37     else{
    38         for(;i<s.length();i++){
    39             cout<<s[i];
    40         }
    41         cout<<endl;
    42     }
    43     return 0;
    44 }
  • 相关阅读:
    DOM
    JavaScript 数组的方法总结
    vuex 状态持久化插件 —— vuex-persistedstate
    移动端1px细线
    CSS多行文本并显示省略号
    Java面试题
    Git提交分支
    Redis的安装配置
    Spring IoC
    单例模式
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4770206.html
Copyright © 2020-2023  润新知