我正在尝试按字典顺序生成给定整数N
编号为K
的分区,例如对于N = 5, K = 3
,我们得到了:
5 = 1 + 1 + 1 + 1 + 1
5 = 1 + 1 + 1 + 2
5 = 1 + 1 + 3
5 = 1 + 2 + 2
5 = 1 + 4
5 = 2 + 3
5 = 5
第三个是1 + 1 + 3
。
如何在不生成每个分区的情况下生成它(在C语言中,但最重要的是我需要算法)?
要在分区中找到最大数字(假设我们可以找到分区数d[i][j]
,其中i
是数字,j
是其分区中的最大整数),然后减少原始分区我们正在寻找的号码和号码。是的,我正在尝试使用动态编程。仍在研究代码。
这根本不起作用:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
FILE *F1, *F2;
main()
{
long long i, j, p, n, k, l, m[102][102];
short c[102];
F1 = fopen("num2part.in", "r");
F2 = fopen ("num2part.out", "w");
n = 0;
fscanf (F1, "%lld %lld", &n, &k);
p = 0;
m[0][0] = 1;
for ( i = 0; i <= n; i++)
{
for (j = 1; j <= i; j++)
{
m[i][j] = m[i - j][j] + m[i][j - 1];
}
for (j = i + 1; j <= n; j++)
{
m[i][j] = m[i][i];
}
}
l = n;
p = n;
j = n;
while (k > 0)
{
while ( k < m[l][j])
{
if (j == 0)
{
while (l > 0)
{
c[p] = 1;
p--;
l--;
}
break;
}
j--;
}
k -=m[l][j];
c[p] = j + 1;
p--;
l -= c[p + 1];
}
//printing answer here, answer is contained in array from c[p] to c[n]
}
答案 0 :(得分:3)
以下是一些生成分区的示例Python代码:
cache = {}
def p3(n,val=1):
"""Returns number of ascending partitions of n if all values are >= val"""
if n==0:
return 1 # No choice in partitioning
key = n,val
if key in cache:
return cache[key]
# Choose next value x
r = sum(p3(n-x,x) for x in xrange(val,n+1))
cache[key]=r
return r
def ascending_partition(n,k):
"""Generate the k lexicographically ordered partition of n into integer parts"""
P = []
val = 1 # All values must be greater than this
while n:
# Choose the next number
for x in xrange(val,n+1):
count = p3(n-x,x)
if k >= count:
# Keep trying to find the correct digit
k -= count
elif count: # Check that there are some valid positions with this digit
# This must be the correct digit for this location
P.append(x)
n -= x
val = x
break
return P
n=5
for k in range(p3(n)):
print k,ascending_partition(n,k)
打印:
0 [1, 1, 1, 1, 1]
1 [1, 1, 1, 2]
2 [1, 1, 3]
3 [1, 2, 2]
4 [1, 4]
5 [2, 3]
6 [5]
这可以用于生成任意分区而不生成所有中间分区。例如,有9253082936723602分区为300.
print ascending_partition(300,10**15)
打印
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 7, 8, 8, 11, 12, 13, 14, 14, 17, 17, 48, 52]
答案 1 :(得分:0)
def _yieldParts(num,lt):
''' It generate a comination set'''
if not num:
yield ()
for i in range(min(num,lt),0,-1):
for parts in _yieldParts(num-i,i):
yield (i,)+parts
def patition(number,kSum,maxIntInTupple):
''' It generates a comination set with sum of kSum is equal to number
maxIntInTupple is for maximum integer can be in tupple'''
for p in _yieldParts(number,maxIntInTupple):
if(len(p) <=kSum):
if(len(p)<kSum):
while len(p) < kSum:
p+=(0,)
print p
patition(40,8,40)
Output:
-------
(40,0,0,0,0,0,0,0)
(39,1,0,0,0,0,0,0)
.
.
.
.