HDU 4593 H - Robot 水题
H - Robot
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87326#problem/H
Description
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
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 <= 10 3)
Output
There are no black lines between cases.
Sample Input
2
1 2 1
1
1 1
Sample Output
1
1
HINT
题意
有n+1个数,有一个数出现了两次,问你这个数是什么
题解:
水题,map搞一搞就好了~
代码:
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 20001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** map<int,int> H;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
H.clear();
int flag=;
for(int i=;i<=n;i++)
{
int x=read();
H[x]++;
if(H[x]==)
{
flag=x;
}
}
printf("%d\n",flag); } }
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