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

题目链接

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

2
6
100

样例输出

Case #1: 5
Case #2: 73

  输出小于等于n的最大supreme number，superme number：该串所有的子序列都为质数。

  注意：子序列（subsequence）可以不连续；子串（substring）必须连续。

  思路：找规律，发现符合要求的数并不多，打表。

  

#include<bits/stdc++.h>
using namespace std;

int num[20]={317,311,173,137,131,113,73,71,53,37,31,23,17,13,11,7,5,3,2,1};

int main()
{
    int T,cas=1;
    cin>>T;
    while(T--)
    {
        string s;
        cin>>s;
        int n=s.length();
        cout<<"Case #"<<cas++<<": ";
        if(n>=4)
            cout<<"317"<<endl;
        else
        {
            int t=0;
            for(int i=0;i<n;i++)
                t=t*10+(s[i]-'0');
            for(int i=0;i<20;i++)
            {
                if(t>=num[i])
                {
                    cout<<num[i]<<endl;
                    break;
                }
            }
        }
    }
    return 0;
}