CF 538B(Quasi Binary-贪心)

最小化求和：用最少量的准二进制数表示整数

最新推荐文章于 2021-08-13 16:54:02 发布

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本博客探讨如何将任意正整数表示为最少数量的准二进制数之和。通过贪心算法策略，确保每个组成数尽可能接近1，以达到最小化目标。实例演示了输入与输出之间的转换过程。

A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.

You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 106).

Output

In the first line print a single integer k — the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.

In the second line print k numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.

Sample test(s)

input
9

output
9
1 1 1 1 1 1 1 1 1 

input
32

output
3
10 11 11 

贪心，每个答案的数尽量取1

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n;
int a[MAXN],siz=0;
int c[MAXN],d[MAXN];
int main()
{
//	freopen("b.in","r",stdin);
//	freopen(".out","w",stdout);
	
	cin>>n;
	d[1]=1;
	Fork(i,2,8) d[i]=d[i-1]*10;
	
	int siz=0;
	while(n) 
	{
		a[++siz]=n%10;
		n/=10;
	}
	
	
	int k=0;
	MEM(c)
	For(i,siz) 
	{
		k=max(k,a[i]);
		For(j,a[i]) c[j]+=d[i];
	}

	cout<<k<<endl<<c[1];
	Fork(i,2,k) cout<<' '<<c[i];cout<<endl;
	
	return 0;
}