2017 ACM暑期多校联合训练

题目链接

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
2
2 3
1 1 1 1
3 3
1 3 3 1

Sample Output
1 2
1 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])

题意：
给定数组b，保存的数组a中的所有子集和的个数，让找个这个数组a，并且按照字典序输出来。

分析：
b数组里面除了第一个元素b[0]肯定为1之外(表示空集的个数有且仅有一个)，其余的第一次出现的b[i]==j（j!=0），那么就表示i这个数字肯定在a数组中存在，而且为第一个并且为最小的元素，同时将这个数的个数减减（相当于减去单独自己本身一个自己的情况），记录下标i。

这样循环的往后加b数组的元素下标偏移i个单位，如果此时两个数的个数均不为0，也就意味这后面的那个数，可以由前面这个数构成，然后让当前下标的数减去b[i],得到的那个数减去b[i]的个数，（相当于减去这个数可以由b[i]组合而成的个数）剩下的肯定就是这个数字本身的个数

就这样循环着往下找

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxm = 1100002;
int b[maxm];
int m;
int a[maxm];

int main()
{
    int t;
    scanf( "%d", &t );
    while( t-- )
    {
        int n;
        scanf( "%d%d", &n, &m );
        for( int i = 0; i <= m; i++ )
        {
            scanf( "%d", b+i );
        }
        b[0] = 0;///空集的情况，不用任何的元素构成，将其赋值为0，不影响后面的计算
        for( int i = 0; i < n; i++ )///循环得到a数组中的第i个元素
        {
            int j;
            for( j = 0; j <= m; j++ )///找到当前的第一个不为0的
            {
                if( b[j] ) break;
            }
            b[j]--;///相当于有一个本身构成的元素
            a[i] = j;///那么就肯定有一个这个元素，而且为当前的最小值
            for( int k = j; k <= m; k++ )///找后面的所有可以由j组合而成的是数
            {
                if( k+j <= m && b[k] && b[k+j] )
                {
                    b[k+j] -= b[k];
                }
            }
        }
        for( int i = 0; i < n; i++ )
        {
            i == 0 ? printf( "%d", a[i] ) : printf( " %d", a[i] );
        }
        printf( "
" );
    }
    return 0;
}