ZOJ.2481 Unique Ascending Array【子集】 2015/09/24

Unique Ascending Array

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given an array of integers A[N], you are asked to decide the shortest array of integers B[M], such that the following two conditions hold.

• For all integers 0 <= i < N, there exists an integer 0 <= j < M, such that A[i] == B[j]
• For all integers 0 =< i < j < M, we have B[i] < B[j]

Notice that for each array A[] a unique array B[] exists.

Input

The input consists of several test cases. For each test case, an integer N (1 <= N <= 100) is given, followed by N integers A[0], A[1], ..., A[N - 1] in a line. A line containing only a zero indicates the end of input.

Output

For each test case in the input, output the array B in one line. There should be exactly one space between the numbers, and there should be no initial or trailing spaces.

Sample Input

8 1 2 3 4 5 6 7 8
8 8 7 6 5 4 3 2 1
8 1 3 2 3 1 2 3 1
0

Sample Output

1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3

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Author: SHI, Xiaohan
Source: Zhejiang Provincial Programming Contest 2005
求数列中出现过的数值，按升序排列，直接暴力过

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int main(){    int n,p[110],a[110],i,j,k;    while( ~scanf("%d",&n),n ){        k = 0;        for( i = 0 ; i < n ; ++i ){            scanf("%d",&p[i]);            bool flag = false;            for( j = 0 ; j < i ; ++j ){                if( p[i] == p[j] ){                    flag = true;                    break;                }            }            if( !flag )                a[k++] = p[i];        }        sort(a,a+k);        for( i = 0 ; i < k ; ++i ){            if( i )                printf(" %d",a[i]);            else printf("%d",a[i]);        }        printf("\n");    }    return 0;}