每日一九度之 题目1041：Simple Sorting

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：4883

解决：1860

题目描述：

You are given an unsorted array of integer numbers. Your task is to sort this array and kill possible duplicated elements occurring in it.

输入：

For each case, the first line of the input contains an integer number N representing the quantity of numbers in this array(1≤N≤1000). Next N lines contain N integer numbers(one number per each line) of the original array.

输出：

For each case ,outtput file should contain at most N numbers sorted in ascending order. Every number in the output file should occur only once.

样例输入：

6
8 8 7 3 7 7

样例输出：

3 7 8

本来想用桶排做的，提交后发现runtime error，应该是数的范围太宽了。

//Asimple
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <queue>
#include <limits.h>
#define INF 0x7fffffff
using namespace std;
const int maxn = 1005;
typedef long long ll;
int n, num;
int a[maxn];

int main(){
    while( ~scanf("%d",&n) ){
        memset(a,0,sizeof(a));
        while( n -- ){
            scanf("%d",&num);
            a[num] = 1;
        }
        int k = 0;
        for(int i=0; i<maxn; i++){
            if( a[i] ){
                printf(k==0?"%d":" %d",i);
                k ++;
            }
        }
        printf("
");
    }
    return 0;
}

正解，用STL中的set正好。简单，明了！！

//Asimple
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <limits.h>
#define INF 0x7fffffff
using namespace std;
const int maxn = 1005;
typedef long long ll;
int n, num;
set<int> s;

int main(){
    while( ~scanf("%d",&n) ){
        s.clear();
        while( n -- ){
            scanf("%d",&num);
            s.insert(num);
        }
        set<int>::iterator it;
        for(it=s.begin(); it!=s.end(); it++){
            printf(it==s.begin()?"%d":" %d",*it);
        }
        printf("
");
    }
    return 0;
}