upc组队赛15 Supreme Number【打表】

Supreme Number

题目链接

题目描述

A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers.

Now lets define a number N as the supreme number if and only if each number made up of an non-empty subsequence of all the numeric digits of N must be either a prime number or 1.

For example, 17 is a supreme number because 1, 7, 17 are all prime numbers or 1, and 19 is not, because 9 is not a prime number.

Now you are given an integer N (2≤N≤10100), could you find the maximal supreme number that does not exceed N?

输入

In the first line, there is an integer T(T≤100000) indicating the numbers of test cases.

In the following T lines, there is an integer N (2≤N≤10100).

输出

For each test case print "Case #x: y", in which x is the order number of the test case and y is the answer.

样例输入

2
6
100

样例输出

Case #1: 5
Case #2: 73

题解

通过打表发现满足条件的最大数为317

坑点在于石油大把10^100写成10100

代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define scac(x) scanf("%c",&x)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pri3(x,y,z) printf("%d %d %d\n",x,y,z)
#define prl(x) printf("%lld\n",x)
#define prl2(x,y) printf("%lld %lld\n",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld\n",x,y,z)
#define ll long long
#define LL long long
#define pb push_back
#define mp make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define MOD 998244353
#define mod 1e9+7
#define N 1000005
const int maxn=10000;
// int prime[1000100]; //打表程序
// bool book[1000100];
// int top;
// void isprime()
// {
//    memset(book, true, sizeof(book));
//    book[1] = false;
//    for(int i=2;i<maxn;i++)
//    {
//      if(book[i])
//      {
//         for(int j=i+i;j<maxn;j+=i)
//         {
//            book[j] = false;
//         }
//      }
//    }
//    for(int i=2;i<=maxn;i++)
//    {
//      if(book[i])
//      {
//         prime[++top] = i;
//      }
//    }
// }
// int main()
// {
//   int n;
//   isprime();
//   book[1] = true;
//   for(int i = 10000; i >= 1000;i--)
//   {
//     if(book[i])
//     {
//       int temp = i;
//       int a = temp%10;
//       temp/=10;
//       int b = temp%10;
//       temp/=10;
//       int c = temp%10;
//       temp /=10;
//       int d = temp;
//       printf("%d %d %d %d\n",d,c,b,a);
//       if(book[a] && book[b] && book[c]  && book[d]  && book[d*10+c] && book[d*10+b] && book[d*10+a] && book[c*10 + b] && book[c*10 + a] && book[b*10 + a] && book[d*100 + c*10 +b] && book[d*100 + c*10 +a] && book[d*100 + b*10 +a] && book[c*100 + b*10 +a])
//       {
//         printf("%d\n",i);
//       }
//     }
//   }
// }
int a[100] = {1,2,3,5,7,11,13,17,23,31,37,53,71,73,113,131,137,173,311,317,10101};
int main()
{
  int n;
  int t;
  sca(t);
  int cas = 1;
  string s;
  while(t--)
  {
      // sca(n);
      n = 0;
      cin>>s;
      if(s.length()>4)
      {
        printf("Case #%d: %d\n",cas++,317);
        continue;
      }
      rep(i,0,s.length())
      {
        n = n*10+s[i]-'0';
      }
      rep(i,0,21)
      {
        if(n<a[i])
        {
          printf("Case #%d: %d\n",cas++,a[i-1]);
          break;
        }
      }
  }
  return 0;
}