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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 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
题目来源
题意：
要找一个不大于n的数 这个数的子序列都是质数或1
思路：
注意， 子序列是先后顺序不变但是不一定连续。子串要求要连续。
所以这些数是有限的，而且个数很少，可以手算的出来
首先一位的只有1,2,3,5,7
2和5只能在最高位
除了1之外其他数只可能最多出现一次
因此，只有20个数：1, 2, 3, 5, 7, 11, 13, 17, 23, 31,37, 53, 71, 73, 113, 131, 137, 173, 311, 317
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x7f7f7f7f7f7f7f7f
using namespace std;
typedef long long LL;
int t;
char s[105];
int nums[100] = {1, 2, 3, 5, 7, 11, 13, 17, 23, 31,
37, 53, 71, 73, 113, 131, 137, 173, 311, 317};
int main()
{
cin >> t;
for (int cas = 1; cas <= t; cas++) {
cin >> s;
if(strlen(s) >= 4){
printf("Case #%d: 317\n", cas);
continue;
}
int n = 0;
for(int i = 0; i < strlen(s); i++){
n *= 10;
n += s[i] - '0';
}
int ans = 0;
for(int i = 0; i < 20; i++){
if(n >= nums[i]){
ans = nums[i];
}
else{
break;
}
}
printf("Case #%d: %d\n", cas, ans);
}
return 0;
}