HDU 5166 Missing number

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Missing number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 78 Accepted Submission(s): 41

Problem Description

There is a permutation without two numbers in it, and now you know what numbers the permutation has. Please find the two numbers it lose.

Input

There is a number T shows there are T test cases below. (T≤10)
For each test case , the first line contains a integers n , which means the number of numbers the permutation has. In following a line , there are n distinct postive integers.(1≤n≤1,000)

Output

For each case output two numbers , small number first.

Sample Input

233 4 511

Sample Output

1 22 3

Source

BestCoder Round #28

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感觉这道题目题意难懂” In following a line , there are n distinct postive integers.(1≤n≤1,000)“这句话的意思是有n个不同的正整数，即有1个2个。。。1000个但没有说明这些正整数的范围。。。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
};
int a[1004];
int b[2];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t,n,mark;
    cin>>t;
    while(t--)
{
mark=0;
cle(a);
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
sort(a+1,a+n+1);
cle(b);
int j=0;
for(int i=n-1;i>=1;i--)
{
if(a[i+1]-a[i]==3)
{
mark=1;cout<<a[i]+1<<" "<<a[i]+2<<endl;break;
}
if(a[i+1]-a[i]==2)
{
b[j++]=a[i]+1;
}
if(a[i+1]-a[i]==1)continue;
}
if(j==2)
{
cout<<b[1]<<" "<<b[0]<<endl;
}
else if(j==0&&!mark)
{
if(a[1]>2)cout<<a[1]-2<<" "<<a[1]-1<<endl;
else cout<<a[n]+1<<" "<<a[n]+2<<endl;
}
else if(j==1)
{
if(a[1]>1)cout<<a[1]-1<<" "<<b[0]<<endl;
else cout<<b[0]<<" "<<a[n]+1<<endl;
}
}
    return 0;
}

标程的意思貌视是n个正整数，每个正整数的范围是1~1000..........不过当时是按照n个大小为整数范围内做的。。

按标程写的解法

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int a[1100];
int n;
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
cin>>t;
while(t--){
cin>>n;
cle(a);
int b;
int Max=-1;
int Min=INF;
for(int i=1;i<=n;i++){
cin>>b;
a[b]++;
Max=max(b,Max);
Min=min(b,Min);
}
int num=0;
vector<int>v;
v.clear();
for(int i=Min;i<=Max;i++){
if(a[i]==0){
v.push_back(i);
num++;
}
}
if(num==2){
cout<<v[0]<<" "<<v[1]<<endl;
}
else if(num==1){
if(Min==1)cout<<v[0]<<" "<<Max+1<<endl;
else cout<<Min-1<<" "<<v[0]<<endl;
}
else if(num==0){
if(Min==1)cout<<Max+1<<" "<<Max+2<<endl;
else if(Min==2)cout<<1<<" "<<Max+1<<endl;
else cout<<Min-2<<" "<<Min-1<<endl;
}
}
    return 0;
}

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