hdu 1492 the numbers of divisors about humble numbers

The number of divisors(約數) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1314    Accepted Submission(s): 637

Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. 

Now given a humble number, please write a program to calculate the number of divisors about this humble number.For examle, 4 is a humble,and it have 3 divisors(1,2,4);12 have 6 divisors.

 

Input

The input consists of multiple test cases. Each test case consists of one humble number n,and n is in the range of 64-bits signed integer. Input is terminated by a value of zero for n.

 

Output

For each test case, output its divisor number, one line per case.

 

Sample Input

4 12 0

 

Sample Output

3 6

 

Author

lcy

 

Source

“2006校園文化活動月”之“校慶杯”大學生程序設計競賽暨杭州電子科技大學第四屆大學生程序設計競賽

 

Recommend

LL

先求出 humble number 的 2，3，5，7 的因數個數，分別爲 twos, threes, fives, sevens，然後從這些因數裏取出若干個，這若干個相乘就得到一個 humble number 的因數。

用生成函數：

2: a = (1 + x + x ^ 2 + ... + x ^ twos)

3: b = (1 + x + x ^ 2 + ... + x ^ threes)

5: c = (1 + x + x ^ 2 + ... + x ^ fives)

7: d = (1 + x + x ^ 2 + ... + x ^ sevens)

然後 a * b * c * d

求 ak * x ^ k 的係數 ak，把所有係數加起來！

#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <algorithm>

const int P_MAX = 100000; // power maximum

int PolyMult (int *a, int pa, int *b, int pb) {
    static int c [P_MAX];
    int pc = pa + pb;
    memset (c, 0, sizeof (*c) * (pc + 1));
    for (int i=0; i<=pa; ++i) {
        for (int k=0; k<=pb; ++k) {
            assert (i + k < P_MAX);
            c [i + k] += a [i] * b [k];
        }
    }
    memcpy (a, c, sizeof (*c) * (pc + 1));
    return pc;
}

int CountFactor (unsigned long long n, int fac) {
    int cnt = 0;
    while (n % fac == 0) {
        n /= fac;
        ++cnt;
    }
    return cnt;
}

int main () {
    static int e [P_MAX];
    std::fill (e, e + P_MAX, 1);

    static int g [P_MAX]; // generator
    unsigned long long n;

#ifdef ONLINE_JUDGE
    while (scanf ("%I64u", &n) == 1 && n) {
#else
    while (scanf ("%llu", &n) == 1 && n) {
#endif
        int twos = CountFactor (n, 2);
        int threes = CountFactor (n, 3);
        int fives = CountFactor (n, 5);
        int sevens = CountFactor (n, 7);

        memset (g, 0, sizeof (*g) * (twos + threes + fives + sevens + 1));
        g [0] = 1;
        int pg = 0; // power of g

        pg = PolyMult (g, pg, e, twos);
        pg = PolyMult (g, pg, e, threes);
        pg = PolyMult (g, pg, e, fives);
        pg = PolyMult (g, pg, e, sevens);

        int sum = 0;
        for (int i=0; i<=pg; ++i) {
            sum += g [i];
        }

        printf ("%d\n", sum);
    }
    return 0;
}