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  • TJU Problem 2857 Digit Sorting

    原题:

    2857.   Digit Sorting
    Time Limit: 1.0 Seconds   Memory Limit: 65536K
    Total Runs: 3234   Accepted Runs: 1704



    Several players play a game. Each player chooses a certain number, writes it down (in decimal notation, without leading zeroes) and sorts the digits of the notation in non-decreasing order, obtaining another number. The player who receives the largest number wins.

    You are given the list of numbers initially chosen by the players. Output the winner's resulting number.

    Input

    The first line of each test case contains an integer N (1 ≤ N ≤ 50), indicating the number of the players. Then N integers followed in the second line. Each number will be between 0 and 100000, inclusive.

    The input is terminated with N = 0.

    Output

    Output one line for each test case, indicating the winner's resulting number.

    Sample Input

    6
    1 10 100 1000 10000 100000
    3
    9638 8210 331
    0
    

    Sample Output

    1
    3689
    



    Source: TJU Team Selection Contest 2007 (1)

     

     

     1 #include <iostream>
     2 #include <algorithm>
     3 #include <string.h>
     4 using namespace std;
     5 
     6 int num[55];
     7 int b[55];
     8 
     9 int main()    {
    10     int N;
    11     memset(b, 0, sizeof(b));
    12     while (cin >> N && N != 0)    {
    13         for (int k = 0; k < N; k++)    {
    14             memset(num, 0, sizeof(num));
    15             int n, count = 0; cin >> n;
    16             for (int i = 0; i < 10; i++)        {
    17                 num[i] = n % 10;
    18                 n /= 10;    count++;
    19                 if (n == 0)    break;
    20             }
    21             sort(num, num + count);
    22             //9for (int i = 0; i < count; i++)    cout << "num["<<i<<"] "<<num[i]<<endl;
    23             //memcpy(b, num, sizeof(b));
    24             //cout << "b["<<k<<"] "<<b[k] << endl;
    25             for (int i = 0; i < count; i++)    {
    26                 if (num[i] != 0)    b[k] = b[k]*10 + num[i];
    27             }
    28             //cout << "b["<<k<<"] "<<b[k] << endl;
    29         }
    30         sort(b, b + N);
    31         cout << b[N - 1] << endl;
    32         memset(b, 0, sizeof(b));
    33     }
    34     return 0;
    35 }

     

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  • 原文地址:https://www.cnblogs.com/QingHuan/p/4263791.html
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