Codeforces Round #315 (Div. 2) B. Inventory
来源:互联网 发布:儿童学编程 编辑:程序博客网 时间:2024/05/22 15:25
B. Inventory
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Sample test(s)
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
题意
给一个长度为
题解
直接开一个
AC代码
#include <cstdio>#include <algorithm>#include <cstring>#include <queue>using namespace std;queue<int> Q;int S[100005];int vis[100005];int n,x;int main(){ scanf("%d",&n); memset(vis,0,sizeof vis); for(int i=0;i<n;i++) { scanf("%d",&S[i]); if(S[i]>n||vis[S[i]]) Q.push(i); else vis[S[i]]=1; } for(int i=1;i<=n;i++) { if(vis[i]==0) { x=Q.front(); Q.pop(); S[x]=i; } } for(int i=0;i<n;i++) printf("%d ",S[i]); return 0;}
- B. Inventory-Codeforces Round #315 (Div. 2)
- Codeforces Round #315 (Div. 2) B. Inventory
- Codeforces Round #315 (Div. 2) B. Inventory
- Codeforces Round #315 (Div. 2)-B. Inventory
- Codeforces Round #315 (Div. 2)569B Inventory(队列)
- Codeforces Round #315 (Div. 2) B. Inventory (STL)
- Codeforces Round #315 (Div. 2) B. Inventory (水题)
- B. Inventory(Codeforces Round #315 (Div. 2)水题)
- Codeforces Round #315 div2 B-Inventory 标记,水题
- Codeforces Round #315 (Div. 2)(A,B)
- Codeforces Round #315 (Div. 2) B 模拟
- Codeforces Round #315 (Div. 1) B
- Codeforces Round #131 (Div. 2) A B
- Codeforces Round #134 (Div. 2)B. Airport
- Codeforces Round #170 (Div. 2) problem B
- Codeforces Round #173 (Div. 2) Problem B
- Codeforces Round #181 (Div. 2) B. Coach
- Codeforces Round #185 (Div. 2)--A,B
- HDU_1558_SegmentSet
- 21-IO流-15-IO流(字符流-缓冲区-复制文本文件)
- 本地推送的使用方法
- hdu 2795(单点修改)
- Unicode
- Codeforces Round #315 (Div. 2) B. Inventory
- 检测CPU利用率的Shell脚本
- hdoj 1863 畅通工程 【最小生成树】
- request.setAttribute()、session.setAttribute()和request.getParameter()的联系与区别
- 算法竞赛入门经典: 第三章 数组和字符串 3.4竖式问题
- POJ 1488 TEX Quotes 串 水
- 21-IO流-16-IO流(字符流-缓冲区-自定义MyBufferedReader-read方法)
- HDOJ 1233 还是畅通工程(最小生成树)
- Num 31 : HDOJ : 1863 畅通工程 [ kruskal( 克鲁斯卡尔 )算法 ] [ 最小生成树 ]