# 8.5 第六场 Yes, Prime Minister

8.5 第六场 Yes, Prime Minister

Time Limit: 10000/10000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 926 Accepted Submission(s): 335

Problem Description

Mr. Hacker’s Department of Administrative Affairs (DAA) has infinite civil servants. Every integer is used as an id number by exactly one civil servant. Mr. Hacker is keen on reducing overmanning in civil service, so he will only keep people with consecutive id numbers in [l,r] and dismiss others.

However, permanent secretary Sir Humphrey’s id number is x and he cannot be kicked out so there must be l≤x≤r. Mr. Hacker wants to be Prime Minister so he demands that the sum of people’s id number ∑ri=li must be a prime number.

You, Bernard, need to make the reduction plan which meets the demands of both bosses. Otherwise, Mr. Hacker or Sir Humphrey will fire you.

Mr. Hacker would be happy to keep as few people as possible. Please calculate the minimum number of people left to meet their requirements.

A prime number p is an integer greater than 1 that has no positive integer divisors other than 1 and p.

Input

The first line contains an integer T(1≤T≤106) - the number of test cases. Then T test cases follow.

The first and only line of each test case contains one integer xi(−107≤xi≤107) - Sir Humphrey’s id number.

Output

For each test case, you need to output the minimal number of people kept if such a plan exists, output −1 otherwise.

Sample Input

10
-2
-1
0
1
2
3
4
5
6
7

Sample Output

6
4
3
2
1
1
2
1
2
1

大概题意：

给出x，求附近最小和为质数的个数。

思路：

坑较多。。。需要多情况考虑，注意负数对应答案，不会有输出-1的情况

代码：

#include<iostream>

using namespace std;
const int N = 21000010;
int primes[N], cnt;     // primes[]存储所有素数
bool st[N];         // st[x]存储x是否被筛掉

void get_primes(int n) {
    st[1] = true;
    st[0] = true;
    for (int i = 2; i <= n; i++) {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++) {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    get_primes(21000000);
    int t, n, k = 0;
    cin >> t;
    while (t--) {
        cin >> n;
        int res = 1;
        if (n < 0) {
            res += 2 * (0 - n) + 1;
            n = 1 - n;
            if (!st[n])
                cout << res << endl;
            else {
                while (st[n] && st[n * 2 + 1]) {
                    n++;
                    res += 2;
                }
                if (!st[n])
                    cout << res << endl;
                else
                    cout << 1 + res << endl;
            }
        } else if (n == 0) {
            cout << 3 << endl;
            continue;
        }
        else if (n == 1) {
            cout << 2 << endl;
            continue;
        }
        else {
            if (!st[n])
                cout << 1 << endl;
            else if ((!st[n * 2 + 1]) || (!st[n * 2 - 1]))
                cout << 2 << endl;
            else {
                res = n * 2 + 1;
                n++;
                res++;
                while (st[n] && st[n * 2 + 1]) {
                    n++;
                    res += 2;
                }
                if (!st[n])
                    cout << res << endl;
                else
                    cout << 1 + res << endl;
            }
        }
    }
    return 0;
}