面试题精选（66）：整数的素数和分解问题

【问题描述】
歌德巴赫猜想说任何一个不小于6的偶数都可以分解为两个奇素数之和。对此问题扩展，如果一个整数能够表示成两个或多个素数之和，则得到一个素数和分解式。对于一个给定的整数，输出所有这种素数和分解式。注意，对于同构的分解只输出一次（比如5只有一个分解2 + 3，而3 + 2是2 + 3的同构分解式）。

例如，对于整数8，可以作为如下三种分解：
(1) 8 = 2 + 2 + 2 + 2
(2) 8 = 2 + 3 + 3
(3) 8 = 3 + 5

【算法分析】
由于要将指定整数N分解为素数之和，则首先需要计算出该整数N内的所有素数，然后递归求解所有素数和分解即可。

C++代码实现如下：
#include <iostream>
#include <vector>
#include <iterator>
#include <cmath>
using namespace std;

// 计算num内的所有素数（不包括num）
void CalcPrimes(int num, vector<int> &primes)
{
    primes.clear();
    if (num <= 2)
        return;
   
    primes.push_back(2);
    for (int i = 3; i < num; i += 2) {
        int root = int(sqrt(i));
        int j = 2;
        for (j = 2; j <= root; ++j) {
            if (i % j == 0)
                break;
        }
        if (j > root)
            primes.push_back(i);
    }
}

// 输出所有素数组合(递归实现)
int PrintCombinations(int num, const vector<int> &primes, int from, vector<int> &numbers)
{
    if (num == 0) {
        cout << "Found: ";
        copy(numbers.begin(), numbers.end(), ostream_iterator<int>(cout, " "));
        cout << '/n';
        return 1;
    }
   
    int count = 0;
   
    // 从第from个素数搜索，从而避免输出同构的多个组合
    int primesNum = primes.size();
    for (int i = from; i < primesNum; ++i) {
        if (num < primes[i])
            break;
        numbers.push_back(primes[i]);
        count += PrintCombinations(num - primes[i], primes, i, numbers);
        numbers.pop_back();
    }
   
    return count;
}

// 计算num的所有素数和分解
int ExpandedGoldbach(int num)
{
    if (num <= 3)
        return 0;
   
    vector<int> primes;
    CalcPrimes(num, primes);
   
    vector<int> numbers;
    return PrintCombinations(num, primes, 0, numbers);
}

int main()
{
    for (int i = 1; i <= 20; ++i) {
        cout << "When i = " << i << ":/n";
        int count = ExpandedGoldbach(i);
        cout << "Total: " << count << "/n/n";
    }
}

运行结果：
When i = 1:
Total: 0

When i = 2:
Total: 0

When i = 3:
Total: 0

When i = 4:
Found: 2 2
Total: 1

When i = 5:
Found: 2 3
Total: 1

When i = 6:
Found: 2 2 2
Found: 3 3
Total: 2

When i = 7:
Found: 2 2 3
Found: 2 5
Total: 2

When i = 8:
Found: 2 2 2 2
Found: 2 3 3
Found: 3 5
Total: 3

When i = 9:
Found: 2 2 2 3
Found: 2 2 5
Found: 2 7
Found: 3 3 3
Total: 4

When i = 10:
Found: 2 2 2 2 2
Found: 2 2 3 3
Found: 2 3 5
Found: 3 7
Found: 5 5
Total: 5

When i = 11:
Found: 2 2 2 2 3
Found: 2 2 2 5
Found: 2 2 7
Found: 2 3 3 3
Found: 3 3 5
Total: 5

When i = 12:
Found: 2 2 2 2 2 2
Found: 2 2 2 3 3
Found: 2 2 3 5
Found: 2 3 7
Found: 2 5 5
Found: 3 3 3 3
Found: 5 7
Total: 7

When i = 13:
Found: 2 2 2 2 2 3
Found: 2 2 2 2 5
Found: 2 2 2 7
Found: 2 2 3 3 3
Found: 2 3 3 5
Found: 2 11
Found: 3 3 7
Found: 3 5 5
Total: 8

When i = 14:
Found: 2 2 2 2 2 2 2
Found: 2 2 2 2 3 3
Found: 2 2 2 3 5
Found: 2 2 3 7
Found: 2 2 5 5
Found: 2 3 3 3 3
Found: 2 5 7
Found: 3 3 3 5
Found: 3 11
Found: 7 7
Total: 10

When i = 15:
Found: 2 2 2 2 2 2 3
Found: 2 2 2 2 2 5
Found: 2 2 2 2 7
Found: 2 2 2 3 3 3
Found: 2 2 3 3 5
Found: 2 2 11
Found: 2 3 3 7
Found: 2 3 5 5
Found: 2 13
Found: 3 3 3 3 3
Found: 3 5 7
Found: 5 5 5
Total: 12

When i = 16:
Found: 2 2 2 2 2 2 2 2
Found: 2 2 2 2 2 3 3
Found: 2 2 2 2 3 5
Found: 2 2 2 3 7
Found: 2 2 2 5 5
Found: 2 2 3 3 3 3
Found: 2 2 5 7
Found: 2 3 3 3 5
Found: 2 3 11
Found: 2 7 7
Found: 3 3 3 7
Found: 3 3 5 5
Found: 3 13
Found: 5 11
Total: 14

When i = 17:
Found: 2 2 2 2 2 2 2 3
Found: 2 2 2 2 2 2 5
Found: 2 2 2 2 2 7
Found: 2 2 2 2 3 3 3
Found: 2 2 2 3 3 5
Found: 2 2 2 11
Found: 2 2 3 3 7
Found: 2 2 3 5 5
Found: 2 2 13
Found: 2 3 3 3 3 3
Found: 2 3 5 7
Found: 2 5 5 5
Found: 3 3 3 3 5
Found: 3 3 11
Found: 3 7 7
Found: 5 5 7
Total: 16

When i = 18:
Found: 2 2 2 2 2 2 2 2 2
Found: 2 2 2 2 2 2 3 3
Found: 2 2 2 2 2 3 5
Found: 2 2 2 2 3 7
Found: 2 2 2 2 5 5
Found: 2 2 2 3 3 3 3
Found: 2 2 2 5 7
Found: 2 2 3 3 3 5
Found: 2 2 3 11
Found: 2 2 7 7
Found: 2 3 3 3 7
Found: 2 3 3 5 5
Found: 2 3 13
Found: 2 5 11
Found: 3 3 3 3 3 3
Found: 3 3 5 7
Found: 3 5 5 5
Found: 5 13
Found: 7 11
Total: 19

When i = 19:
Found: 2 2 2 2 2 2 2 2 3
Found: 2 2 2 2 2 2 2 5
Found: 2 2 2 2 2 2 7
Found: 2 2 2 2 2 3 3 3
Found: 2 2 2 2 3 3 5
Found: 2 2 2 2 11
Found: 2 2 2 3 3 7
Found: 2 2 2 3 5 5
Found: 2 2 2 13
Found: 2 2 3 3 3 3 3
Found: 2 2 3 5 7
Found: 2 2 5 5 5
Found: 2 3 3 3 3 5
Found: 2 3 3 11
Found: 2 3 7 7
Found: 2 5 5 7
Found: 2 17
Found: 3 3 3 3 7
Found: 3 3 3 5 5
Found: 3 3 13
Found: 3 5 11
Found: 5 7 7
Total: 22

When i = 20:
Found: 2 2 2 2 2 2 2 2 2 2
Found: 2 2 2 2 2 2 2 3 3
Found: 2 2 2 2 2 2 3 5
Found: 2 2 2 2 2 3 7
Found: 2 2 2 2 2 5 5
Found: 2 2 2 2 3 3 3 3
Found: 2 2 2 2 5 7
Found: 2 2 2 3 3 3 5
Found: 2 2 2 3 11
Found: 2 2 2 7 7
Found: 2 2 3 3 3 7
Found: 2 2 3 3 5 5
Found: 2 2 3 13
Found: 2 2 5 11
Found: 2 3 3 3 3 3 3
Found: 2 3 3 5 7
Found: 2 3 5 5 5
Found: 2 5 13
Found: 2 7 11
Found: 3 3 3 3 3 5
Found: 3 3 3 11
Found: 3 3 7 7
Found: 3 5 5 7
Found: 3 17
Found: 5 5 5 5
Found: 7 13
Total: 26

修改上面的程序，去掉分解式的输出语句，只保留计数功能，分别对N = 100, 200, 300进行测试。结果如下：
-bash-3.2$ time ./a.out
When i = 100:
Total: 40899

real    0m0.114s
user    0m0.112s
sys     0m0.002s

-bash-3.2$ time ./a.out
When i = 200:
Total: 9845164

real    0m38.344s
user    0m38.329s
sys     0m0.001s

-bash-3.2$ time ./a.out
When i = 300:
Total: 627307270

real    50m37.198s
user    50m32.556s
sys     0m1.786s
-bash-3.2$

由结果可见，当N = 300时，分解数就大得吓人了。