题目链接：https://nanti.jisuanke.com/t/31452

A prime number (or a prime) is a natural number greater than $1$ that cannot be formed by multiplying two smaller natural numbers.

Now lets define a number $N$ as the supreme number if and only if each number made up of an non-empty subsequence of all the numeric digits of $N$ must be either a prime number or $1$.

For example, $17$ is a supreme number because $1$, $7$, $17$ are all prime numbers or $1$, and $19$ is not, because $9$ is not a prime number.

Now you are given an integer N (2≤N≤1e100), could you find the maximal supreme number that does not exceed $N$?

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

In the following T lines, there is an integer N (2≤N≤10100).

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

注意：

子序列（Subsequence）和子串（Substring）是不一样的，子序列可以是不连续的。

题意：

若一个数的所有非空子序列是素数（或者 $1$），则称它为“supreme number”，

现在给出一个 $N$，要求不大于 $N$ 的最大的supreme number。

题解：

考虑五位数：首先由于2,5,7只要有两个同时出现，就不行，所以2,5,7只能挑一个；又3不能出现超过1次，所以只能要一个3；那么剩下3个空位只能填1，但一旦有111就不是素数，就不行。

所以超过四位的数统统不行，则只要暴力把四位以内的数全部找出来即可。

AC代码：

#include<bits/stdc++.h>

using namespace std;

int n[]={,,,,,,,,,,,,,,,,,,,,};

char str[];

int main()

{

    int T;

    cin>>T;

    for(int kase=;kase<=T;kase++)

    {

        scanf("%s",str);

        int len=strlen(str);

        if(len>=) printf("Case #%d: %d\n",kase,n[]);

        else

        {

            int num=,k=;

            for(int i=len-;i>=;i--)

            {

                num+=(str[i]-'')*k;

                k*=;

            }

            printf("Case #%d: %d\n",kase,n[upper_bound(n+,n+,num)-n-]);

        }

    }

}