沈阳网络赛K-Supreme Number【规律】

•  26.89%
•  1000ms
•  131072K

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

ACM-ICPC 2018 沈阳赛区网络预赛

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
", 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
", cas, ans);
	}
	return 0;
}