确定的是多个可能的组合的数量来获得指定结果 [英] Determining the number of possible combinations of a number to get a specified result
问题描述
我遇到了这样一个问题:
给定的整数,确定可能的组合仅使用2,3,7-数,其总和会给出整数
例如:
4 - 2 {(2,2)}
9 - 3 - {(2,7),(2,2,2,3),(3,3,3)}
一种方法是遍历3圈,然后再确定和是否是可以实现的。这里的code:
为(i = 0; I< = NUM / 2;我++){
为(J = 0; J< = NUM / 3; J ++){
对于(K = 0; K< = NUM / 7; k ++){
如果(I * 2 + J * 3 + K * 7 = = NUM)
算上++;
}
下面计数将具有套可能的数目。但是,这是非常低效和需要O(N)的时间。我想知道是否有计算不同套数的任何其他有效的方式。
一个DP的解决方案应该为这个问题应该是线性的。 (这里执行)
的#include< stdio.h中>
#定义SZ 5
INT备忘录[SZ + 1 + 7]。
诠释的主要(无效){
INT I = 0;
memset的(安培;备忘录[0],0,sizeof的备忘录);
备忘录[0] = 1;
对于(i = 0; I< = SZ; ++ I)备注[I + 2] + =备忘录[I]
对于(i = 0; I< = SZ; ++ I)备注[I + 3] + =备忘录[I]
对于(i = 0; I< = SZ; ++ I)备注[I + 7 + =备忘录[I]
的printf(%D \ N,备忘录[深圳]);
返回0;
}
- 我们先从一维数组DP
备忘录
与理想infinited大小 不引起出大小的(在实际中动态分配) 开往指数SZ + MAX_NUM
。 - 在用来初始化元素
0
这个数组1,因为有1路 获得empty_sum。 - 如果我们能得到一些
K
x中的方式,有x的方式来 获得K + 2
,K + 3
和K + 7
。这是3环被使用。 (Number_of_ways [{2,3,7} + 1] + = number_of_ways [I]) - 在所有循环完成后,备忘录[K:0 - SZ]包含的数 方法我们可以得到氏/ STRONG>
给人一种复杂性为O(K * N),其中k为3这里(2,3,7)。为常数k,这是线性的。
I came across this question:
Given an integer, determine the number of possible combinations using only 2,3,7 whose sum will give the integer.
Eg:
4 - 2 {(2,2)}
9 - 3 {(2, 7), (2, 2, 2, 3), (3, 3, 3)}
One way is to iterate through 3 loops and then determine whether the sum is attainable. Here's the code:
for( i=0; i<=num/2; i++){
for( j=0; j<=num/3; j++){
for( k=0; k<=num/7; k++){
if(i*2+j*3+k*7 == num)
count++;
}
Here count will have the number of sets possible. But this is very inefficient and takes O(n3) time. I would like to know if there is any other efficient way of computing the number of different sets.
A dp solution should for this problem should be linear. (Implemented here)
#include <stdio.h>
#define SZ 5
int memo[SZ+1+7];
int main(void) {
int i = 0;
memset(&memo[0], 0, sizeof memo);
memo[0] = 1;
for(i = 0; i <= SZ; ++i) memo[i+2] += memo[i];
for(i = 0; i <= SZ; ++i) memo[i+3] += memo[i];
for(i = 0; i <= SZ; ++i) memo[i+7] += memo[i];
printf("%d\n", memo[SZ]);
return 0;
}
- We start with a 1-D dp array
memo
with ideally infinited size (allocated dynamically in practice) of size which doesn't cause out of bound for indexSZ + max_num
. - Initilize element
0
of this array with 1, because there is 1 way to obtain empty_sum. - If we can obtain a number
k
in x ways, there are x more ways to obtaink+2
,k+3
andk+7
. This is what 3 loops are using. (Number_of_ways[{2,3,7}+i] += number_of_ways[i]) - After all loops are done, memo[k : 0 - SZ] contains the number of ways we can obtain k.
Giving a complexity of O(k * N) where k is 3 here (2, 3, 7). For constant k, this is linear.
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