Supreme Number

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本文介绍了一种特殊的数——Supreme数，这类数的所有非空子序列组成的数字要么为质数要么为1。文章探讨了如何找出不超过给定整数的最大Supreme数，并提供了解决方案的代码实现。

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A prime number (or a prime) is a natural number greater than 11 that cannot be formed by multiplying two smaller natural numbers.

Now lets define a number NN as the supreme number if and only if each number made up of an non-empty subsequence of all the numeric digits of NN must be either a prime number or 11.

For example, 1717 is a supreme number because 11, 77, 1717 are all prime numbers or 11, and 1919 is not, because 99 is not a prime number.

Now you are given an integer N\ (2 \leq N \leq 10^{100})N (2≤N≤10100), could you find the maximal supreme number that does not exceed NN?

Input

In the first line, there is an integer T\ (T \leq 100000)T (T≤100000) indicating the numbers of test cases.

In the following TT lines, there is an integer N\ (2 \leq N \leq 10^{100})N (2≤N≤10100).

Output

For each test case print "Case #x: y", in which xxis the order number of the test case and yy is the answer.

样例输入

2
6
100

样例输出

Case #1: 5
Case #2: 73

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题意：一个supreme数是其的所有子集都是质数或者1（可以是不连续的子集），求小于n的最大的supreme数

~一言难尽，当时以为连续呢,(Q_Q)

思路：考虑单独的一个数是supreme数或1的有1,2,3,5,7，二位数是supreme数的有73,71,53,37,31,23,17,13,11，三位数是supreme数的有317,311,173,137,131,113，四位数没有supreme数

代码：

#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
char s[1100];
int a[20]={1,2,3,5,7,11,13,17,23,31,37,53,71,73,113,131,137,173,311,317};
int main()
{
    ll t,cas=1;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%s",s);
        int l=strlen(s);
        printf("Case #%lld: ",cas++);
        if(l>=4)
          printf("317\n");
        else
        {
            int f=0,i=0;
            while(s[i])
              f=f*10+(s[i]-'0'),i++;
            for(i=19;i>=0;i--)
            {
                if(a[i]<=f)
                {
                   printf("%d\n",a[i]);
                   break;
                }
            }
        }
    }
    return 0;
}