PAT（甲级）2019年秋季考试 7-1 Forever (20 分)

“Forever number” is a positive integer $A$ with $K$ digits, satisfying the following constrains:

• the sum of all the digits of $A$ is $m$;
• the sum of all the digits of $A+1$ is $n$; and
• the greatest common divisor of $m$ and $n$ is a prime number which is greater than 2.

Now you are supposed to find these forever numbers.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integer $N$ ($\le 5$). Then $N$ lines follow, each gives a pair of $K$ ($3<K<10$) and $m$ ($1<m<90$), of which the meanings are given in the problem description.

Output Specification:

For each pair of $K$ and $m$, first print in a line Case X, where X is the case index (starts from 1). Then print $n$ and $A$ in the following line. The numbers must be separated by a space. If the solution is not unique, output in the ascending order of $n$. If still not unique, output in the ascending order of $A$. If there is no solution, output No Solution.

Sample Input:

2
6 45
7 80

Sample Output:

Case 1
10 189999
10 279999
10 369999
10 459999
10 549999
10 639999
10 729999
10 819999
10 909999
Case 2
No Solution

题意

给出K和m。计算所有满足条件的数A，A有K位，A的数字和是m，A+1的数字和n与m的最大公约数是大于2的质数。输出满足条件的n和A，按照升序排列，其中n是第一标尺，A是第二标尺。

思路

暴力的话会超时，可以dfs并且剪枝，剪枝的依据是当前的数字和是否还有可能产生满足条件的A。
话说这道题放到第1题的位置真的好吗……

代码

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

bool isPrime(int x) {
    if (x <= 1) return false;
    for (int i = 2, sqr = sqrt(x); i <= sqr; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}

int gcd(int a, int b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

int digitSum(int x) {
    int sum = 0;

    string s = to_string(x);
    for (int i = 0, len = s.length(); i < len; ++i)
        sum += s[i] - '0';

    return sum;
}

struct record {
    int sum, val;

    record(int v, int n) : val(v), sum(n) {}

    bool operator<(record &x) {
        if (sum != x.sum) return sum < x.sum;
        else return val < x.val;
    }
};

vector<record> r;

void dfs(int sum, int val, int left, int target) {
    if (left == 0 && sum == target) {
        int n = digitSum(val + 1), g = gcd(sum, n);
        if (g > 2 && isPrime(g)) r.push_back(record(val, n));
    } else if (left > 0)
        for (int i = 0; i <= 9; ++i)
            if (sum + i + left * 9 - 9 >= target && sum + i <= target)
                dfs(sum + i, val * 10 + i, left - 1, target);
}

int main() {
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);

    int n, k, m;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        r.clear();
        cin >> k >> m;
        cout << "Case " << i << "\n";
        for (int j = 1; j <= 9; ++j) dfs(j, j, k - 1, m);
        if (r.empty()) cout << "No Solution\n";
        else {
            sort(r.begin(), r.end());
            for (int j = 0; j < r.size(); ++j)
                cout << r[j].sum << " " << r[j].val << "\n";
        }
    }
}