line rate input fin one urn scanf inpu p s

Numbers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 514 Accepted Submission(s): 270

Problem Description zk has n numbers a1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj). These new numbers could make up a new sequence b1，b2,...,bn(n−1)/2

.
LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can‘t figure out which numbers were in a or b. "I‘m angry!", says zk.
Can you help zk find out which n numbers were originally in a?

Input Multiple test cases(not exceed 10).
For each test case:
?The first line is an integer m(0≤m≤125250), indicating the total length of a and b. It‘s guaranteed m can be formed as n(n+1)/2.
?The second line contains m numbers, indicating the mixed sequence of a and b.
Each ai is in [1,10^9]

Output For each test case, output two lines.
The first line is an integer n, indicating the length of sequence a;
The second line should contain n space-seprated integers a1,a2,...,an(a1≤a2≤...≤an). These are numbers in sequence a.
It‘s guaranteed that there is only one solution for each case. 題意 已經有an個數 對於 1≤i<j≤n，an中的數 兩兩相加，得到b數組 然後把a，b數組亂序 讓你挑選出a數組的內容

#include<bits/stdc++.h>
using namespace std;
const int maxn = 125250+100;
map<int,int>mp;
int s[maxn];//記錄a和b
int t[maxn];//存儲a數組的結果

void init()
{
    mp.clear();
    memset(t,0,sizeof(t));
}

int main()
{
    int n;
    while(~scanf("%d",&n) )
    {
        init();
        for(int i=1;i<=n;i++)
            scanf("%d",&s[i]);
        sort(s+1,s+n+1);
        int tot = 0;
        for(int i=1;i<=n;i++)
        {
            if(tot + (tot-1)*tot/2 >= n)//總數夠了  就不需要再選下去了
                break;
            int x= s[i];
            if(mp[x]==0 || mp.count(x)==0)//前面是這個數在mp中 且數值為1 另外一個是這個不在mp中
            {
                for(int j=0;j<tot;j++)
                {
                    mp[x+t[j]]++;//把新挑出來的數和之前的都相加
                }
                t[tot++] = x;//存儲這個數
            }
            else
            {
                mp[x]--;//這個數之前出現過  所以不需要加了
            }
        }
        cout<<tot<<endl;
        for(int i=0;i<tot;i++)
        {
            if(i)
                cout<<" ";
            cout<<t[i];
        }
        cout<<endl;
    }
    return 0;
}

hdu 6168 Numbers