此代码是由某人设计的,用于将数组[a1 a2...am b1 b2..bn ]
更改为数组[b1 b2 ..bn a1 a2..am]
,但它涉及最大的公约数,我无法理解这一点。
void Exchange(int a[],int m,int n,int s){
int p=m,temp=m+n;int k=s%p;
while(k!=0){temp=p;p=k;k=temp%p;}
for(k=0 ; k<p ;k++){ //below is where i cant't understand
temp=a[k];i=k;j=(i+m)%(m+n);
while(j!=k)
{a[i]=a[j];i=j;j=(j+m)%(m+n);}
a[i]=temp;
}
};
编辑:“正确”缩进:
void Exchange(int a[], int m, int n, int s) {
int p = m, temp = m + n, k = s % p;
while (k != 0) {
temp = p;
p = k;
k = temp % p;
}
for (k = 0 ; k < p; k ++) { // below is where i cant't understand
temp = a[k];
i = k;
j = (i + m) % (m + n);
while (j != k) {
a[i] = a[j];
i = j;
j = (j + m) % (m + n);
}
a[i] = temp;
}
};
答案 0 :(得分:0)
代码使用单个开销值来实现数组旋转。如果长度是相互素数,则单次通过就足够了。如果没有,则必须通过长度为
的GCD重复移位循环我之前说过,还有其他问题可以解决这个问题。找到SO 3333-3814的外观处理单个旋转。我做了一些乱码代码以支持前一段时间,证明需要GCD,但我以前没有发布过它。
这是代码 - 它使用C99 VLAs - 可变长度数组。
#include <stdio.h>
static int gcd(int x, int y)
{
int r;
if (x <= 0 || y <= 0)
return(0);
while ((r = x % y) != 0)
{
x = y;
y = r;
}
return(y);
}
static void dump_matrix(int m, int n, int source[m][n])
{
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
printf("%4d", source[i][j]);
putchar('\n');
}
}
static void init_matrix(int m, int n, int source[m][n])
{
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
source[i][j] = (i + 1) * (j + 2);
}
}
static void rotate_1col(int n, int vector[n], int z)
{
z %= n;
if (z != 0)
{
int c = gcd(n, z);
int s = n / c;
for (int r = 0; r < c; r++)
{
int x = r;
int t = vector[x];
for (int i = 0; i < s; i++)
{
int j = (x + z) % n;
int v = vector[j];
vector[j] = t;
x = j;
t = v;
}
}
}
}
static void rotate_cols(int m, int n, int source[m][n], int z)
{
for (int i = 0; i < m; i++)
rotate_1col(n, source[i], z);
}
int main(void)
{
int m = 3;
for (int n = 2; n < 9; n++)
{
int source[m][n];
for (int z = 0; z <= n; z++)
{
init_matrix(m, n, source);
printf("Initial:\n");
dump_matrix(m, n, source);
rotate_cols(m, n, source, z);
printf("Post-rotate %d:\n", z);
dump_matrix(m, n, source);
putchar('\n');
}
}
return 0;
}
代码演示了不同大小的数组的不同旋转大小。输出的示例部分:
…
Initial:
2 3 4
4 6 8
6 9 12
Post-rotate 1:
4 2 3
8 4 6
12 6 9
…
Initial:
2 3 4 5
4 6 8 10
6 9 12 15
Post-rotate 3:
3 4 5 2
6 8 10 4
9 12 15 6
…
Initial:
2 3 4 5 6 7
4 6 8 10 12 14
6 9 12 15 18 21
Post-rotate 1:
7 2 3 4 5 6
14 4 6 8 10 12
21 6 9 12 15 18
Initial:
2 3 4 5 6 7
4 6 8 10 12 14
6 9 12 15 18 21
Post-rotate 2:
6 7 2 3 4 5
12 14 4 6 8 10
18 21 6 9 12 15
Initial:
2 3 4 5 6 7
4 6 8 10 12 14
6 9 12 15 18 21
Post-rotate 3:
5 6 7 2 3 4
10 12 14 4 6 8
15 18 21 6 9 12
…
Initial:
2 3 4 5 6 7 8 9
4 6 8 10 12 14 16 18
6 9 12 15 18 21 24 27
Post-rotate 4:
6 7 8 9 2 3 4 5
12 14 16 18 4 6 8 10
18 21 24 27 6 9 12 15
Initial:
2 3 4 5 6 7 8 9
4 6 8 10 12 14 16 18
6 9 12 15 18 21 24 27
Post-rotate 5:
5 6 7 8 9 2 3 4
10 12 14 16 18 4 6 8
15 18 21 24 27 6 9 12
Initial:
2 3 4 5 6 7 8 9
4 6 8 10 12 14 16 18
6 9 12 15 18 21 24 27
Post-rotate 6:
4 5 6 7 8 9 2 3
8 10 12 14 16 18 4 6
12 15 18 21 24 27 6 9
…
答案 1 :(得分:0)
首先,为了得到你所说的预期结果,我将m和n设置为数组大小的一半。我还假设s将初始化为零,在这种情况下,第一个while循环不会迭代。此外,您的代码中缺少多个声明,因此我的解释会做出一些假设。
The variable p holds the number of array elements to swap;
// This is to keep the value to be overwritten by the swap
temp=a[k];
// This is the array index of the bottom half element to write the top half element to
i=k;
// this is to get the current index of the top half;
j=(i+m)%(m+n);
// This assignes the bottom index value with the top half value
while(j!=k)
{
// Write top half element to corresponding bottom half element
a[i]=a[j];
// We can now overwrite top half element; this assignes the index at wich to copy the bottom half element
i=j;
// This is to get out of the loop
j=(j+m)%(m+n);
}
// The bottom half element held at the beginning is now written to the top half at the corresponding index
a[i]=temp;
希望这是您正在寻找的答案。我通过使用调试器并逐行踩入代码来达到这个结果。我不知道你是否知道如何使用调试器,但如果没有,那么我强烈推荐你的精益如何使用它;它花了很多时间,并返回了一个很棒的红利: - )