Divisibility by Thirty-six(字符串维护+整除问题+贪心)

最新推荐文章于 2020-04-11 15:33:45 发布

原创最新推荐文章于 2020-04-11 15:33:45 发布 · 345 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#思维

练习专栏收录该内容

160 篇文章

订阅专栏

本文介绍了一种算法，该算法接收一个由0到9的数字字符串组成的数据集，长度在1到1000之间，目标是利用这些数字构建能够被36整除的最大数值。每个数字只能使用一次，输出结果为该最大数值的十进制表示形式，如果无法构造则输出'不可能'。

Description

There is a string containing only decimal digit characters. The length of the string is between 1 and 1,000. Using characters of the string, you have to construct the maximum number which divides by thirty-six without remainder. Each character of the string may not be used more than once.

Input

The first line of the input contains an integer T (T <= 1000), indicating the number of cases. Each case begins with a line containing a single line representing the source string.

Output

For each test case, print a line containing the test case number (beginning with 1) and the decimal representation of the maximum number (leading zeroes should be omitted). If no number can be constructed, please output “impossible” instead.

Sample Input

2
03061
345

Sample Output

Case 1: 6300
Case 2: impossible

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 1005
#define M 15
int dir[30][2]={{0, 0},{4, 0},{8, 0},{1, 2},{1, 6},{2, 0},{2, 4},{2, 8},{3, 2},{3, 6},{4, 4},{8, 4},{5, 2},{5, 6},{6, 0},{6, 4},{6, 8},{7, 2},{7, 6},{8, 8},{9, 2},{9, 6}};
int s[N],temp,rec[M],cnt[M],flag,id,len;
char ans[N];
void put(int n[],int k,char *str)
{
    int l=0,i,j;
    for(i=9;i>=0;i--)
    {
        for(j=0;j<n[i];j++)
        {
            str[l]=i+'0';
            l+=1;
        }
    }
    for(i=0;i<2;i++)
    {
        str[l]=dir[k][i]+'0';
        l+=1;
    }
    str[l]='\0';
}
int compare(char *str,char *s)
{
    int i,j;
    for(i=0;str[i]!='\0';i++)
    {
        if(str[i]!=s[i]) return str[i]<s[i];
    }
    return 0;
}
void judge(int n)
{
    char now[N];
    put(cnt,id,now);
    if(n<len||(compare(ans,now)&&n==len))
    {
        flag=1;
        len=n;
        strcpy(ans,now);
    }
}
void dfs(int n,int d,int sum)
{
    if(d>=n)
    {
        if((sum-temp)%9==0) judge(n);
            return ;
    }
    for(int i=1;i<=8;i++)
    {
        if(!cnt[i]) continue;
        cnt[i]-=1;
        dfs(n,d+1,sum+i);
        cnt[i]+=1;
    }
}
void Search(int d)
{
    for(int i=0;i<2;i++)
    {
        cnt[dir[d][i]]-=1;
        if(cnt[dir[d][i]]<0) return ;
    }
    id=d;
    for(int i=0;i<=temp&&i<=len;i++)
    {
        dfs(i,0,0);
    }
}
int isok(char *str)
{
    int sum=0,l=strlen(str);
    if(l==0&&rec[0]==0) return 0;
    for(int i=0;i<l;i++)
    {
        sum+=str[i]-'0';
    }
    return !sum;
}
void solve()
{
    int i;
    for(i=0;i<22;i++)
    {
        memcpy(cnt,rec,sizeof(rec));
        Search(i);
    }
}
int main()
{
   int T,kase=1;
   scanf("%d",&T);
   while(T--)
   {
      char num[N];
      scanf("%s",num);
      len=strlen(num);
      temp=flag=0;
      memset(rec,0,sizeof(rec));
      memset(ans,0,sizeof(ans));
      for(int i=0;i<len;i++)
      {
          int item=num[i]-'0';
          temp+=item;
          rec[item]+=1;
      }
      solve();
      if(isok(ans))
         printf("Case %d: 0\n",kase);
      else if(flag)
         printf("Case %d: %s\n",kase,ans);
      else
         printf("Case %d: impossible\n",kase);
      kase+=1;
   }
   return 0;
}