Integer division recursive algorithm, and all the divided output (Java)
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Algorithm is still working, this is an integer divide problem.
The Internet for a long time, most are seeking integer division of the number of species, rarely seeking detailed breakdown situation.
There are also C language version, and can not find a Java.
So I came up with a Java (almost the same, but if someone needs Java version of the easy to search)
problem
Explain the problem, what is the integer division?
n = m1 + m2 + ... + mi; (where mi is a positive integer and 1 <= mi <= n), then the {m1, m2, ..., mi} is a division of n.
If {m1, m2, ..., mi} is not more than the maximum value m, i.e., max (m1, m2, ..., mi) <= m, it is said to belong to a m n divided. Here we denote the number n is divided by m f (n, m);
For example, when n = 6 we can get that several of the following division:
6 = 6
6 5 + 1 =
6 2 = 4 +
6 + 4 = 1 + 1
6 3 + 3 =
6 + 2 + 1 = 3
6 3 + 1 = 1 + 1 +
6 2 + 2 + 2 =
6 = 2 + 2 + 1 + 1
6 + 1 2 = 1 + 1 + 1 +
6 = 1 + 1 + 1 + 1 + 1 + 1
analysis
This way it is very easy to see, n for an integer division is based on n-1 based on, so that there may be recurrence formula.
Difficult to find a single parameter recursive, so we add a parameter m, a division result represents the largest addend.
The relationship between n and m, consider the following cases:
-
When n = 1, regardless of the value of the number m (m> 0), i.e., only one partitioning {1};
-
When m = 1, n has a value regardless of the number, n, of only one division 1, {1,1,1, ..., 1};
-
When n = m, divided according to whether to include n, it can be divided into two situations:
(A) a case containing n division, i.e., only a {n};
(B) n does not include the division, the largest number of time division certainly smaller than n, i.e. all n (n-1) division. Therefore q (n, n) = 1 + q (n, n-1);
-
When n <m, since the negative division impossible, therefore equivalent to q (n, n);
-
However, n> m, the maximum value according to division contains m, it can be divided into two situations:
(A) contained in the case of division of m, i.e., nm {m, {x1, x2, ... xi}}, where {x1, x2, ... xi} and is, therefore this case is q (nm, m)
The case (b) does not contain m division, the division of all values smaller than m, i.e., the n (m-1) is divided, the number of q (n, m-1);
Therefore q (n, m) = q (nm, m) + q (n, m-1);
Code
Seeking integer division species
/*
* @Title division
* @Description 整数划分方法
* @author 滑技工厂
* @Date 2020/3/8
* @param [n, m] n->要划分的整数 m->最大加数
* @return int 输出有多少种划分
* @throws
*/
public static int division(int n, int m) {
if (n < 1 || m < 1)
return 0;
if (n == 1 || m == 1)
return 1;
//最大加数如果大于n 则另最大加数为n
if (n < m)
return division(n, n);
//m=n，则从
if (n == m)
return division(n, m - 1) + 1;
return division(n, m - 1) + division(n - m, m);
}
Seeking specific division of
static int mark[] = new int[100];//记录分解情况
static int n;
/*
* @Title divide
* @Description 输出划分
* @author 滑技工厂
* @Date 2020/3/8
* @param [now, k, pre]
* @return void
* @throws
*/
public static void divide(int now, int k, int pre) {
int i;
//数组长度大于n就返回
if (now > n) return;
if (now == n) {
System.out.printf("%d=", n);
for (i = 0; i < k - 1; i++) {
System.out.printf("%d+", mark[i]);
}
System.out.printf("%d\n", mark[i]);
} else {
for (i = pre; i > 0; i--) {
if (i <= pre) {
mark[k] = i;
now += i;
divide(now, k + 1, i);
now -= i;
}
}
}
}
Results 6 as an example, is obtained
End Sahua ✿, gifts of roses.
