计蒜客31452 Supreme Number（找规律)

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 xx is the order number of the test case and yy is the answer.

样例输入复制

2
6
100

样例输出复制

Case #1: 5
Case #2: 73

题意：求出最大的不超过n的由素数组成(所有数字组合情况都是素数)的元素
分析：
    1、数字一定由1，2，3，5，7组成
    2、数字中2，5，7两两之间不能同时出现（25，52，27，72，57，75都不是素数），由此可以推断出全部符合的数都  不超过四位
    2、除1以外都不能重复出现两次和以上（会被11整除）例：313中33被11整除

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
	int T,cnt=1;
	scanf("%d",&T);
	int a[20]={1,2,3,5,7,11,13,17,23,31,37,53,71,73,113,131,137,173,311,317};
	while(T--)
	{
		int n=0;
		char s[111];
		scanf("%s",s);
		printf("Case #%d: ",cnt++);
		if(strlen(s)>=4)
			printf("317
");
		else
		{
			for(int i=0;s[i];i++)
				n=n*10+s[i]-'0';
			for(int i=19;i>=0;i--)
				if(a[i]<=n){
					printf("%d
",a[i]);
					break;
				}
		}
	}
	return 0;
}