Rikka with Subset HDU - 6092 （DP+组合数）

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 nn positive A1−AnA1−An and their sum is mm. Then for each subset SS of AA, Yuta calculates the sum of SS. 

Now, Yuta has got 2n2n numbers between [0,m][0,m]. For each i∈[0,m]i∈[0,m], he counts the number of iis he got as BiBi. 

Yuta shows Rikka the array BiBi and he wants Rikka to restore A1−AnA1−An. 

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

InputThe first line contains a number t(1≤t≤70)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)n,m(1≤n≤50,1≤m≤104).

The second line contains m+1m+1 numbers B0−Bm(0≤Bi≤2n)B0−Bm(0≤Bi≤2n).OutputFor each testcase, print a single line with nn numbers A1−AnA1−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:一个含有n个数的数组，他们的sum和是m，并且这n个数的所有子集的sum和的个数用一个数组b来表示。
其中b[i] 表示 子集中sum和为i的有b[i]个。现在给你N，M，b数组，让你推出数组a，并输出字典序最小的那一个。

思路：可以知道满足条件的只有一个数集set，所以只需要排序输出就是字典序最小的那个了。
那么如何求数组a呢？，首先我们应该知道，如果有n个0，会产生2^n个sum和为0的集合。
那么数组a中0的数量直接就是log2(b[0])了。
而sum和为1的只需要用所以的sum和为0的集合加上一个1即可，
所以num[1] = b[1]/b[0];
然后定义数组dp[i]，表示不用数字i，仅用小于i中的数凑出来sum和为i的集合数量。
那么（b[i]-dp[i]）/b[0] 就是Num[i]
求dp[i]的过程中用到dp的思想，细节见代码。

    if (dp[j] == 0) continue;
    if (num[i] == 0) break;

这步的代码可以节省时间复杂度。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 100010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int t;
int n, m;
ll b[maxn];
int dp[maxn];
int num[maxn];
int c(int n, int m)
{
    int sum = 1;
    for (int i = n - m + 1; i <= n; i++) sum *= i;
    for (int i = 1; i <= m; i++) sum /= i;
    return sum;
}
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    // gbtb;
    // cin >> t;
    scanf("%d", &t);
    while (t--)
    {
        MS0(num);
        MS0(dp);
        scanf("%d%d", &n, &m);
        repd(i, 0, m)
        {
            scanf("%lld", &b[i]);
            // cin >> b[i];
        }
        num[0] = log2(b[0]);
        num[1] = b[1] / b[0];
        dp[0] = b[0];
        repd(i, 0, m)
        {
            for (int j = m; j >= 0; j--)
            {
                if (dp[j] == 0) continue;
                if (num[i] == 0) break;
                for (int k = 1; k <= num[i]; k++)
                {
                    if (j + k*i <= m)
                    {
                        dp[j + k*i] += dp[j] * c(num[i], k);
                    }
                }
            }
                if (i + 1 <= m)
                {
                    num[i+1]=(b[i+1]-dp[i+1])/b[0];
                }
        }
        bool flag = 0;
        for (int i = 0; i <= m; i++)
        {
            for (int j = 1; j <= num[i]; j++)
            {
                if (!flag) printf("%d", i), flag = 1;
                else printf(" %d", i);
            }
        }
        puts("");
    }



    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}