HDU 6168 Numbers【水题】

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 356    Accepted Submission(s): 180

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

62 2 2 4 4 4211 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

 

Sample Output

32 2 261 2 3 4 5 6

 

Source

2017 Multi-University Training Contest - Team 9

题意：

给你两个序列A,B，序列B中的数是A中任意两个数的和。现在给出你A,B序列混在一起的数，让你找出A序列输出。

思路：

每次加进去一个筛掉后最小的数，然后与前面已经有的数把剩下的数筛一下就好。

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1.0);const int mod = 1e9 + 7;const int maxn = 125350;int a[maxn], ans[maxn];int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);int n;while (~scanf("%d", &n)){map<int, int> mp;for (int i = 0; i < n; i++){scanf("%d", a + i);mp[a[i]]++;}int num = 0;for (int i = 0; i < n; i++){if (mp[a[i]])//加进去与前面有的筛一下就好{mp[a[i]]--;ans[num++] = a[i];for (int k = num - 2; k >= 0; k--)if (mp[ans[num - 1] + ans[k]])mp[ans[num - 1] + ans[k]]--;}}printf("%d\n", num);for (int i = 0; i < num; i++){cout << ans[i];if (i != num - 1) printf(" ");}puts("");}return 0;}