contest: Codeforces Round #300, problem: (B) Quasi Binary

B. Quasi Binary

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

91 1 1 1 1 1 1 1 1 

input

32

output

310 11 11 

思路：例如543=500+40+3   表示500最少需要5个数，表示40最少需要4个数，表示3最少需要3个数。

最大的数字为5，就一定至少要用5个数来表示。

因为可以输出任一组符合要求的数 所以输出 111 111 111 110 100

用字符数组做 15ms

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    char a[10],b[10][10];    while(~scanf("%s",a))    {        memset(b,0,sizeof(b));        int cnt=0;        int len=strlen(a);        for(int i=0;i<len;i++)           {                int m=a[i]-'0';                if(m>cnt) cnt=m;                for(int j=0;j<m;j++)                    b[j][i]='1';           }        printf("%d\n",cnt);        for(int i=0;i<cnt;i++)        {            int flag=0;            for(int j=0;j<len;j++)              if(b[i][j]=='1') flag=1;              else if(b[i][j]!='1'&&flag)  b[i][j]='0';   ///所缺的0，不是前导0        }        for(int i=0;i<cnt;i++)        {            for(int j=0;j<len;j++)               if(b[i][j]=='0'||b[i][j]=='1') printf("%c",b[i][j]);   ///除了1 0 以外的都不需要输出            if(i==cnt-1) puts("");            else  putchar(' ');        }    }    return 0;}

用int做  15ms

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){   int n;   int num[7]={1,10,100,1000,10000,100000,1000000};   while(~scanf("%d",&n))   {       int a[7],b[10];       memset(a,0,sizeof(a));         memset(b,0,sizeof(b));      ///b记录要输出的每一组数       int cnt=0;       int ans=0;       while(n)       {        a[cnt++]=n%10;        if(n%10>ans) ans=n%10;        n/=10;       }             <span style="font-family: Arial, Helvetica, sans-serif;">///找出最大的数字</span>       printf("%d\n",ans);          for(int j=0;j<cnt;j++)       {          for(int i=0;i<a[j];i++)                       b[i]+=num[j];                 ///比较容易错...       }                         for(int i=0;i<ans-1;i++)       printf("%d ",b[i]);       printf("%d\n",b[ans-1]);   }   return 0;}