ACM-ICPC 2018 沈阳赛区网络预赛 K Supreme Number

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

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int sum[] = {
1,
2,
3,
5,
7,
11,
13,
17,
23,
31,
37,
53,
71,
73,
113,
131,
137,
173,
311,
317,
48481548
};
const int binggg=317;
string s;
int toint(const string &s) {
	stringstream ss;
	ss<<s;
	int res;
	ss>>res;
	return res;
}

int solve(int n) {
	if(n>=binggg) return binggg;
	for(int i=0; ;i++) {
		if(sum[i+1]>n) return sum[i];
	}
}

int main() {
	int T;
	int kase=1;
	cin>>T;
	while(T--) {
		cin>>s;
		cout<<"Case #"<<kase++<<": ";
		if(s.size()>5) cout<<binggg<<endl;
		else {
			int n=toint(s);
			cout<<solve(n)<<endl;
		}
	}
}