hdu 多校联赛 6092 Rikka with Subset

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta calculates the sum of S. 

Now, Yuta has got 2n numbers between [0,m]. For each i∈[0,m], he counts the number of is he got as Bi.

Yuta shows Rikka the array Bi and he wants Rikka to restore A1−An.

It is too difficult for Rikka. Can you help her?  

 

Input

The first line contains a number t(1≤t≤70), the number of the testcases. 

For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104).

The second line contains m+1 numbers B0−Bm(0≤Bi≤2n).

 

Output

For each testcase, print a single line with n numbers A1−An.

It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.

 

Sample Input

22 31 1 1 13 31 3 3 1

 

Sample Output

1 21 1 1
Hint
In the first sample, $A$ is $[1,2]$. $A$ has four subsets $[],[1],[2],[1,2]$ and the sums of each subset are $0,1,2,3$. So $B=[1,1,1,1]$
 

 

Source

2017 Multi-University Training Contest - Team 5

题意：比赛的时候就没看懂题意 赛后补题也没看懂 感觉题目之间量的关系都描述的不清晰 根据多方查询+百度+瞎猜才明白了个大体意思 题目先说了有A这个数组 数组有n个数 n已知 然后这个数组能形成2^n个子集（子集的解释看样例）这2的n次方个A的子集的和的大小为i  i在（0，m）内  其中m是确定已知的 然后这些子集和为i的数量又形成了数组B 数组B也已知 现在然你回过头来求A（感觉真的是够绕的……）

解析：很多个较小的数字随机组合会求出多个很大的数字，所以从B0向Bm推导，在每求出A序列的一部分这个过程中，更新后续的B序列，更新完的B[i]就是 i 在A序列中出现的次数。 
分析完后，主要的难点就是怎么去让已求出来的A序列随机组合，更新后续的B序列直接减就可以了。看成01背包问题，让m为背包去装 i，初始值为dp[0] = 1，由于i依次增大，A子集随机组合不会重复

ac代码：

#include <bits/stdc++.h>#define N 10010using namespace std;int a[N], b[N], dp[N],c[N];//dp[i]表示：加和为i的子集个数；int main(){    int t, n, m;    scanf("%d", &t);    while(t--)    {//先清空所有数组        memset(a, 0, sizeof(a));//用来存储求得的a数组的值        memset(b, 0, sizeof(b));//存储输入的到的b数组的值        memset(dp, 0, sizeof(dp));//用来存储dp过程中规划的过程值        scanf("%d %d", &n, &m);        for(int i = 0; i <= m; i++)            scanf("%d", &b[i]);//读取b数组        dp[0] = 1;//初始化规划数组的初值        int p = 0;//中间变量 做a数组的下标        for(int i = 1; i <= m; i++)        {            c[i] = b[i] - dp[i];//c数组存储A序列中值为i的个数            for(int j = 0; j < c[i]; j++)            {                a[p++] = i; //对A序列赋值                for(int k = m; k>= i; k--) //处理成01背包                {                    dp[k] += dp[k - i]; //和为k的A子集个数相加去更新B序列                }            }        }        for(int i = 0; i < p; i++)        {            if(i > 0) printf(" ");//注意规范输出格式 防止re            printf("%d", a[i]);//输出A序列        }        printf("\n");    }    return 0;}