题目链接：https://ac.nowcoder.com/acm/contest/984/G/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

The cows are forever competing to see who has the best brand. The latest rage is brands that are 'Wonderprimes'. You probably already know that a brand consists of a sequence of digits that does not begin with 0; brands actually look a lot like positive integers.
A wonderprime is a number that can be partitioned into two prime numbers, each of which has at least D digits and, of course, doesn't start with 0. When D=2, the number 11329 is a wonderprime (since it connects 113 and 29, both of which are prime).
Only a few of the cows have wonderprime brands, but they all want one. Your job is to find the first wonderprime greater than or equal to a supplied integer N (1 <= N <= 2,000,000,000). No integer greater than 2,000,000,000 will be required.

输入描述

Line 1: Two space-separated integers: D and N

输出描述

Line 1: A line that contains the first wonderprime no smaller than N.

输入

2 11328

输出

11329

解题思路

题意:找到一个大于等于n且由都不小于d位的两个素数组成的一个数。
思路:直接暴力，从n开始，一个一个的试。

Accepted Code:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll d, n, mod = 10, minimum = 1;
bool prime(ll x) {
    if (x < 2)
        return false;
    for (ll i = 2; i * i <= x; i++)
        if (x % i == 0)
            return false;
    return true;
}
bool iskey(ll n) {
    ll x = n / mod, y = n % mod, pow = mod;
    if (n % mod / minimum)
        if (prime(x) && prime(y))
            return true;
    y += x % 10 * pow;
    x /= 10;
    while (x >= minimum) {
        if (prime(x) && prime(y))
            return true;
        pow *= 10;
        y += x % 10 * pow;
        x /= 10;
    }
    return false;
}
int main() {
    scanf("%lld%lld", &d, &n);
    for (ll i = 1; i < d; i++) {
        mod *= 10;
        minimum *= 10;
    }
    if (n < mod * minimum)
        n = mod * minimum;
    while (!iskey(n))
        n++;
    printf("%lld\n", n);
    return 0;
}