使用它们的数组成员 [英] array members working with them
问题描述
嗨。
i的代码如下：
Hi.
i have the code below:
int dynamicbinomial(int n, int k)
{
int* b = new int[n,k];
for(int i=0; i<= n; i++)
{
for(int j=0; j<= min(i,k); j++)
{
if(j==0 || j==i)
b[i,j] = 1;
else if(j==1 || j==i-1)
b[i,j] = i;
else
b[i,j] = b[i-1,j-1] + b[i-1,j];
}
}
return b[n,k];
}
它工作正常，但在最后一行：
it works correctly but in the last line:
b[i,j] = b[i-1,j-1] + b[i-1,j];
到达该行时它确实如此：
b [i，j] = b [0，j-1] + b [1，j-1] + b [2，j-1] + b [3， j-1] + ... + b [i-1，j];
它显示行（i-1）中所有数字的总和0到j
如果你输入5作为n和3作为k你必须有一个矩阵如下：
when it reaches that line it does like it:
b[i,j] = b[0,j-1] + b[1,j-1] + b[2,j-1] + b[3,j-1] + ... + b[i-1,j] ;
it shows the sum of all numbers in line (i-1) form 0 to j
if you enter 5 as n and 3 as k you must have a matrix as below:
1
1 1
1 2 1
1 3 3
1 4 6
1 5 10
但它会生成一个类似矩阵：
1
1 1
1 2 1
1 3 3
1 4 7
1 5 12
跟踪它：
1- i = 0 j = 0 k = 3 min（i，k）= 0 b [i，j] = b [0,0] = 1
2- i = 1 j = 0 k = 3 min（i，k）= 1 b [i，j] = b [1,0] = 1
3 i = 1 j = 1 k = 3 min（i，k）= 1 b [i，j] = b [1,1] = 1
4- i = 2 j = 0 k = 3 min（i，k）= 2 b [i，j] = b [2,0] = 1
5 i = 2 j = 1 k = 3 min（i，k）= 2 b [i，j] = b [2,1] = 2
6- i = 2 j = 2 k = 3 min（i，k）= 2 b [i，j] = b [2,2] = 1
7- i = 3 j = 0 k = 3 min（i，k）= 3 b [i，j] = b [3,0] = 1
8- i = 3 j = 1 k = 3 min（i，k）= 3 b [i，j] = b [3,1] = 3
9- i = 3 j = 2 k = 3 min（i，k）= 3 b [i，j] = b [3,2] = 3
10- i = 3 j = 3 k = 3 min（i，k）= 3 b [i，j] = b [3,3] = 1
11- i = 4 j = 0 k = 3 min（i，k）= 3 b [i，j] = b [4,0] = 1
12- i = 4 j = 1 k = 3 min（i，k）= 3 b [i，j] = b [4,1] = 4
13- i = 4 j = 2 k = 3 min（i，k）= 3 b [i，j] = b [4,2] = 7
1 4- i = 4 j = 3 k = 3 min（i，k）= 3 b [i，j] = b [4,3] = 4
15- i = 5 j = 0 k = 3 min（i，k）= 3 b [i，j] = b [5,0] = 1
16- i = 5 j = 1 k = 3 min（i，k）= 3 b [i，j] = b [5,1] = 5
17- i = 5 j = 2 k = 3 min（i，k）= 3 b [i，j] = b [5,2] = 12
见第13行？
but it makes a matrix like it:
1
1 1
1 2 1
1 3 3
1 4 7
1 5 12
traced it this way:
1- i= 0 j= 0 k= 3 min( i,k )= 0 b[i,j]= b[0,0]= 1
2- i= 1 j= 0 k= 3 min( i,k )= 1 b[i,j]= b[1,0]= 1
3- i= 1 j= 1 k= 3 min( i,k )= 1 b[i,j]= b[1,1]= 1
4- i= 2 j= 0 k= 3 min( i,k )= 2 b[i,j]= b[2,0]= 1
5- i= 2 j= 1 k= 3 min( i,k )= 2 b[i,j]= b[2,1]= 2
6- i= 2 j= 2 k= 3 min( i,k )= 2 b[i,j]= b[2,2]= 1
7- i= 3 j= 0 k= 3 min( i,k )= 3 b[i,j]= b[3,0]= 1
8- i= 3 j= 1 k= 3 min( i,k )= 3 b[i,j]= b[3,1]= 3
9- i= 3 j= 2 k= 3 min( i,k )= 3 b[i,j]= b[3,2]= 3
10- i= 3 j= 3 k= 3 min( i,k )= 3 b[i,j]= b[3,3]= 1
11- i= 4 j= 0 k= 3 min( i,k )= 3 b[i,j]= b[4,0]= 1
12- i= 4 j= 1 k= 3 min( i,k )= 3 b[i,j]= b[4,1]= 4
13- i= 4 j= 2 k= 3 min( i,k )= 3 b[i,j]= b[4,2]= 7
14- i= 4 j= 3 k= 3 min( i,k )= 3 b[i,j]= b[4,3]= 4
15- i= 5 j= 0 k= 3 min( i,k )= 3 b[i,j]= b[5,0]= 1
16- i= 5 j= 1 k= 3 min( i,k )= 3 b[i,j]= b[5,1]= 5
17- i= 5 j= 2 k= 3 min( i,k )= 3 b[i,j]= b[5,2]= 12
see line 13 ?
推荐答案
我修改了你的功能。希望能帮助到你。确保释放分配的内存。
I have modified your function. Hope it helps. Make sure to release the allocated memory.
int dynamicbinomial(int n, int k)
{
int **b = new int*[n];
for(int i = 0; i < n; ++i) {
b[i] = new int[k];
}
// Print the address of the allocated array
for (int i=0; i<n; ++i)
{
for (int j=0; j<k; j++)
{
std::cout<< &b[i][j] << " ";
}
std::cout<<"\n";
}
for(int i=0; i< n; i++)
{
for(int j=0; j< std::min(i+1,k); j++)
{
if(j==0 || j==i)
b[i][j] = 1;
else if(j==1 || j==i-1)
b[i][j] = i;
else
{
b[i][j] = b[i-1][j-1] + b[i-1][j];
}
// Print the values and address
std::cout<<b[i][j] << " " <<std::cout<< &b[i][j] << " ";
}
std::cout<<"\n";
}
return b[n][k];
}
您是否仔细查看过您的代码？
让我们稍微缩小一下......
Have you looked closely at your code?
Let's just shrink it a bit...
int* b = new int[n, k];
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= Math.Min(i, k); j++)
{
b[i, j] = 1;
}
}数组中有多少个元素，有多少元素分配给1？
how many elements are there in the array, and how many are you assigning "1" to?
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