hdu6092 Rikka with Subset Rikka with Subset

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概述

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 997 Accepted Submission(s): 493

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

i s 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

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 $B=[1,1,1,1]$

题解：

如果 $B_i$ 是 $B$ 数组中除了 $B_0$ 以外第一个值不为 $0$ 的位置，那么显然 $i$ 就是 $A$ 中的最小数。

现在需要求出删掉 $i$ 后的 $B$ 数组，过程大概是反向的背包，即从小到大让 $B_j-=B_{j-i}$ 。

时间复杂度 $O (n m)$ 。

我写了两个代码，思想本质上是一样的

#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define ll long long
#define read(a) scanf("%d",&a);
using namespace std;
const int maxn=10005;
ll a[maxn],b[maxn];
int main(){
	freopen("test.txt","r",stdin);
	int t;
	int n,m;
	scanf("%d",&t);
	while(t--){
		scanf("%d %d",&n,&m);
		for(int i=0;i<=m;i++){
			scanf("%lld",&b[i]);
		}
		int cnt=0;
		int i=1;
		while(i<=m){
			while(b[i]>0){
				a[++cnt]=i;
				b[i]--;
				for(int j=i+1;j<=m;j++){
					if(b[j]>0)
						b[j]-=b[j-i];
				}
			}
			i++;
		}
		for(i=1;i<=cnt;i++){
			if(i>1)
				printf(" ");
			printf("%lld",a[i]);
		}
		printf("n");
	}
	return 0;
}

#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define ll long long
//#define read(a) scanf("%d",&a);
using namespace std;
const int maxn=10005;
ll a[maxn],b[maxn];
ll dp[maxn];
int main(){
	freopen("test.txt","r",stdin);
	int t;
	int n,m;
	scanf("%d",&t);
	while(t--){
		memset(dp,0,sizeof(dp));
		scanf("%d %d",&n,&m);
		for(int i=0;i<=m;i++){
			scanf("%lld",&b[i]);
		}
		int cnt=0;
		dp[0]=1;
		for(int i=1;i<=	m;i++){
			b[i]-=dp[i];
			for(int j=1;j<=b[i];j++){
				a[++cnt]=i;
				for(int k=m;k>=i;k--){
					dp[k]+=dp[k-i];
				}
			}
			//i++;
		}
		for(int i=1;i<=cnt;i++){
			if(i>1)
				printf(" ");
			printf("%lld",a[i]);
		}
		printf("n");
	}
	return 0;
}