UVA - 10905 - Children's Game (简单排序)
UVA - 10905
Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
4thIIUCInter-University Programming Contest, 2005 |
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A |
Children’s Game |
Input: standard input |
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Problemsetter: Md. Kamruzzaman |
There are lots of number games for children. These games are pretty easy to play but not so easy to make. We will discuss about an interesting game here. Each player will be given Npositive integer. (S)He can make a big integer by appending those integers after one another. Such as if there are 4 integers as 123, 124, 56, 90 then the following integers can be made – 1231245690, 1241235690, 5612312490, 9012312456, 9056124123 etc. In fact 24 such integers can be made. But one thing is sure that 9056124123 is the largest possible integer which can be made.
You may think that it’s very easy to find out the answer but will it be easy for a child who has just got the idea of number?
Input
Each input starts with a positive integer N (≤ 50). In next lines there are N positive integers. Input is terminated by N = 0, which should not be processed.
Output
For each input set, you have to print the largest possible integer which can be made by appending all the N integers.
Sample Input |
Output for Sample Input |
4 |
9056124123 |
Source
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Data Structures and Libraries :: Linear Data Structures with Built-in Libraries :: C++ STL algorithm (Java Collections)
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: General Problem Solving Techniques ::Exercises: Intermediate
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design
Root :: Prominent Problemsetters :: Md. Kamruzzaman (KZaman)
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Data Structures and Libraries :: Linear Data Structures with Built-in Libraries :: C++ STL algorithm (Java Collections)
找对排序方法就好了!
AC代码:
#include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <iostream> using namespace std; string num[55]; bool cmp(string a, string b) { return a+b > b+a; //‘+’表示a和b的连接 } int main() { int n; while(scanf("%d", &n), n) { for(int i=0; i<n; i++) { cin >> num[i]; } sort(num, num+n, cmp); for(int i=0; i<n; i++) { cout << num[i]; } cout << endl; } return 0; }
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