hdu--6168--Numbers

最新推荐文章于 2018-10-16 16:20:10 发布

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本文介绍了一种解决混合序列问题的方法，通过分析已知的混合数组找出原始序列。该问题涉及一系列数学操作及排序算法的应用，对于理解序列处理及算法实现具有一定帮助。

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Numbers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 162 Accepted Submission(s): 86

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：

有n个数的序列a（a1......an），可以两两组合成n（n-1）／2个数的b序列。把它们混成一个数组，让你求出原n个数的a序列。

解题思路：

因为b序列是由两个a序列的元素组合而成，所以混合的这个序列按从小到大排序后，前两个较小的数一定为a序列中的元素。假设为z[1],z[2],那么z[1]+z[2]就一定存在于b数组中。把z[1]+z[2]用map“标记”一下，继续从z[3]遍历混合的数组，判断z[3]有没有被“标记”，如果有说明这个数为b序列中的数，跳过继续遍历下一个。如果没有，该数就为a数组中的元素。以此类推.........

代码：

 C++ Code 

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#include <bits/stdc++.h>
using namespace std;
const int N = 125250 + 8;
map<int, int> m;
int a[N], z[N];
int main()
{
    int n, k;
    while(~scanf("%d", &n))
    {
        k = 0;
        m.clear();
        for(int i = 1; i <= n; i++)
            scanf("%d", &z[i]);
        sort(z + 1, z + 1 + n);
        a[k++] = z[1];
        a[k++] = z[2];
        m[z[1] + z[2]]++; ///表示b数组中包含该数
        for(int i = 3; i <= n; i++)
        {
            if(m[z[i]] > 0)
            {
                ///如果b数组中有这个数，
                m[z[i]]--;
                continue;
            }
            else
            {
                a[k++] = z[i];
                for(int j = 0; j < k - 1; j++)
                    m[a[j] + z[i]]++;
            }
        }
        sort(a, a + k);
        printf("%d\n", k);
        for(int i = 0; i < k; i++)
        {
            if(i)
                printf(" ");
            printf("%d", a[i]);
        }
        printf("\n");
    }
}