zoukankan      html  css  js  c++  java
  • CodeForces

    思路:这道题要重复数字的下标存起来,把没出现过的数字标记起来并且存起来,然后再把重复的数字换成没出现过的(没出现过的数字的数量 = 重复数字的数量)

    Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.

    During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.

    You have been given information on current inventory numbers for n items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to n by changing the number of as few items as possible. Let us remind you that a set of n numbers forms a permutation if all the numbers are in the range from 1 to n, and no two numbers are equal.

    Input

    The first line contains a single integer n — the number of items (1 ≤ n ≤ 105).

    The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the initial inventory numbers of the items.

    Output

    Print n numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.

    Examples

    Input
    3
    1 3 2
    Output
    1 3 2 
    Input
    4
    2 2 3 3
    Output
    2 1 3 4 
    Input
    1
    2
    Output
    1 

    Note

    In the first test the numeration is already a permutation, so there is no need to change anything.

    In the second test there are two pairs of equal numbers, in each pair you need to replace one number.

    In the third test you need to replace 2 by 1, as the numbering should start from one.

    #include <cstdio>
    #include <iostream>
    #include <cmath>
    #include <string>
    #include <cstring>
    #include <queue>
    using namespace std;
    int a[100000+8], sign[100000+8], tag[100000+8], position[100000+8];
    int main()
    {
        int n, miao = 0, root = 0;
        memset(sign, 0, sizeof(sign));
        scanf("%d", &n);
        for(int i = 0; i<n; i++)
        {
            scanf("%d", &a[i]);
            if(!sign[a[i]] && a[i] <= n && a[i]) sign[a[i]] = 1;
            else tag[miao++] = i;//把重复出现的数字的下标存起来
        }
        for(int i = 1; i <= n; i++)
            if(!sign[i]) position[root++] = i;//把没出现过的数字存起来
        for(int i = 0; i<miao; i++)//重复数字的个数与没出现过的数字的个数相等
        {
            a[tag[i]] = position[i];//把重复出现的数字的位置换上没出现过的最小的数字
        }
        int flag = 0;//输出
        for(int i = 0; i<n; i++)
        {
            if(flag)printf(" ");
            flag = 1;
            printf("%d", a[i]);
        }
        printf(" ");
        return 0;
    }
  • 相关阅读:
    python简单的游戏场景代码
    ZooKeeper应用场景
    Zookeeper配置文件中的配置项解释和Zookeeper的安装
    MapReduce的手机流量统计的案例
    java和js获取当前天之后或之前7天(任意)日期
    JDK中ThreadDump诊断Java代码中的线程死锁问题
    Java代码调用Oracle的存储过程,存储函数和包
    Oracle的卸载过程步骤
    JDK中ConcurrentHashMap效率测试
    JDK提供的四种线程池代码详解
  • 原文地址:https://www.cnblogs.com/RootVount/p/10472945.html
Copyright © 2011-2022 走看看