hdu 4593 Robot

文章描述了Glad Corporation公司发现其生产的机器人编号中有一个编号被重复使用的问题,并提供了一个算法来找出这个重复的编号。

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=4593  

Robot

Description

A robot is a mechanical or virtual artificial agent, usually an electro-mechanical machine that is guided by a computer program or electronic circuitry. Robots can be autonomous or semi-autonomous and range from humanoids such as Honda's Advanced Step in Innovative Mobility (ASIMO) and Tosy's TOSY Ping Pong Playing Robot (TOPIO) to industrial robots, collectively programmed 'swarm' robots, and even microscopic nano robots. By mimicking a lifelike appearance or automating movements, a robot may convey a sense of intelligence or thought of its own.
Robots have replaced humans in the assistance of performing those repetitive and dangerous tasks which humans prefer not to do, or are unable to do due to size limitations, or even those such as in outer space or at the bottom of the sea where humans could not survive the extreme environments.
After many years, robots have become very intellective and popular. Glad Corporation is a big company that produces service robots. In order to guarantee the safety of production, each robot has an unique number (each number is selected from 1 to N and will be recorded when the robot is produced).
But one day we found that N+1 robots have been produced in the range of 1 to N , that's to say one number has been used for 2 times. Now the president of Glad Corporation hopes to find the reused number as soon as possible.

Input

Multiple cases, end with EOF.
In each case, The first line has one number N, which represents the maximum number. The next line has N +1 numbers. (All numbers are between 1 to N, and only two of them are the same.) (1 <= N <= 103)

Output

Each case, output a line with the reused number.
There are no black lines between cases.

Sample Input

2
1 2 1
1
1 1

Sample Output

1
1

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::map;
using std::min;
using std::find;
using std::pair;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1010;
const int INF = 0x3f3f3f3f;
map<int, int> A;
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w+", stdout);
#endif
    int n, v, ans;;
    while(~scanf("%d", &n)) {
        A.clear();
        rep(i, n + 1) {
            scanf("%d", &v);
            A[v]++;
            if(2 == A[v]) ans = v;
        }
        printf("%d\n", ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4792975.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值