2017 ACM暑期多校联合训练

题目链接

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.

Sample Input
6
2 2 2 4 4 4
21
1 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

Sample Output
3
2 2 2
6
1 2 3 4 5 6

题意：

已知数组a由n也元素组成，数字b由数组a里面的任意两个元素相加组成，则数组b里面的元素个数为n(n-1)/2.

现在a数组和b数组混合在一起一共有m个元素，求解a数组里面的元素个数，并将这些元素打印出来。

分析：

首先可以确定的是，混合数组里面的两个最小的元素一定是a数组里面的值，这样每次将找到的值与a数组中已有的值相加得到的值从混合数组里面删去，这样混合数组中现存的第一个值肯定就是要加入到a数组里面的，接着重复进行上面的过程，直到找齐a数组里面的元素。

其实就是个存粹的模拟的过程。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4;
typedef long long ll;

ll a[maxn];
int main()
{
    ll num, m;
    while(~scanf("%lld", &m))
    {
        map <ll, ll> M;
        ll min1=1000000009,min2=1000000000;///数列中的次小值和最小值
        ll n=(sqrt(8*m+1)-1)/2;///根据n与m的关系，列出n的表达式
        for(ll i=0; i<m; i++)
        {
            scanf("%lld",&num);
            if(num<min2)///如果比最小值还小，两个值都要更新
            {
                min1=min2;
                min2=num;
            }
            else if(num<min1)///比次小值小，只更新次小值就行了
            {
                min1=num;
            }
            M[num]++;
        }
        M[min1] --;
        M[min2] --;
        a[0] = min2;
        a[1] = min1;
        ll ans = 1;
        while(ans != n - 1)
        {
            for(ll i = 0; i < ans; ++ i)
            {
                M[a[i] + a[ans]] --;
            }
            map<ll, ll>::iterator it = M.begin();
            while(it!= M.end())///找到map容器里面第一个非0 的元素
            {
                if(it -> second == ll(0))
                {
                    M.erase(it ++);
                }
                else
                {
                    break;
                }
            }
            a[++ ans] = M.begin() -> first;///这个值肯定是a里面的一个元素
            M[a[ans]] --;
        }
        printf("%lld
",n);
        printf("%lld",a[0]);
        for(int i=1; i<n; i++)
            printf("% lld",a[i]);
        printf("
");
    }
    return 0;
}