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
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;
}
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