hdu 6092 递推

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 108    Accepted Submission(s): 41

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：

a数组中有n个数，任意组合得到2的n次方个数字，给出这些数字的出现的频率b数组，求原数组a

题解：

从小到大枚举，我们很容易知道第一个数，那么就可以通过第一个数任意组合推知接下来的第二个

比赛的时候是这样的思路，但是循环写的不一样没有break，导致超时，调换循环结构然后剔除不必要的就而已过了

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAXN 10005long long b[MAXN],t[MAXN];int ans[MAXN];long long c[30][30];void init(){    memset(c,0,sizeof(c));    for(int i=0; i<=21; i++)    {        c[i][0]=1;        for(int j=1; j<=i; j++)            c[i][j]=c[i-1][j-1]+c[i-1][j];    }}int main(){    init();    int n,m,T;    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(int i=0; i<=m; i++)            scanf("%lld",&b[i]);        int i,j,k;        for(i=1;i<=m;i++){            if(b[i]==0)                continue;            for(j=i+1;j<=m;j++){                for(k=1;k<=b[i];k++){                    if(j-i*k<i)   break;                    if(j-i*k!=i)  b[j]-=c[b[i]][k]*b[j-i*k];                    else          b[j]-=c[b[i]][k+1];                }            }        }        int flag=1;        for(i=1;i<=m;i++)        {            while(b[i]--)            {                if(flag) flag=0;                else  printf(" ");                printf("%d", i);            }        }        puts("");    }    return 0;}