Supreme Number 2018沈阳icpc网络赛 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 xxis 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：一个supreme数是其的所有子集都是质数或者1（可以是不连续的子集），求小于n的最大的supreme数

分析：考虑单独的一个数是supreme数或1的有1,2,3,5,7，二位数是supreme数的有73,71,53,37,31,23,17,13,11，三位数是supreme数的有317,311,173,137,131,113，四位数没有supreme数

所以我们直接用个数组保存下来这些supreme数，然后看输入的数处于哪个范围就行

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 1e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll prime[] = {317,311,173,137,131,113,73,71,53,37,31,23,17,13,11,7,5,3,2,1};
int main() {
    ll T;
    cin >> T;
    for( ll cas = 1; cas <= T; cas ++ ) {
        string s;
        cin >> s;
        cout << "Case #" << cas << ": ";
        if( s.length() >= 4 ) {
            cout << 317 << endl;
        } else {
            ll sum = 0;
            for( ll i = 0; i < s.length(); i ++ ) {
                sum = sum*10 + (s[i]-'0');
            }
            for( ll i = 0; i < 20; i ++ ) {
                if( sum >= prime[i] ) {
                    cout << prime[i] << endl;
                    break;
                }
            }
        }
    }
    return 0;
}