输出混乱 [英] Confusion on output
问题描述
大家好，
我在以下问题上做了解决方案，但没有想要的输出是
实现。如果你知道某种方式，请告诉我它的评论。
---------------------------------------- -------------
（声明一个二维数组a的大小为10乘以10和
将10个整数声明为[0] [0]到[0] [9]。然后重复由整数序列指定的
插入9次，并且
存储[a]的结果1]到a [9]。最后，打印阵列。
输出示例：
1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3）
-------------------------------- -----------------------
我做了：
#include< stdio.h>
int main（）
{
int a [10] [10] ;
int i，j，k;
int x;
for（i = 0; I< 10; i ++）{
for（j = 0; j< 10; j ++）a [i] [j] = j + 1;
}
for（i = 0; i< 10; i ++）{
for（j = 0; j< i + 1; j ++）{
x = a [i] [j];
a [i] [j] = a [i] [j];
a [i] [j] = x;
for（k = 0; k <10; k ++）printf（"％d"，a [j] [k]）;
printf（ " \ n"）;
}
printf（" \ n"）;
}
返回0;
}
但输出完全不同。
tanx提前
hi all,
I did the solution on the following question,but no desired output is
acheived.if you know some way on,please show me with its comments.
-----------------------------------------------------
(Declare a two-dimensional array"a" of size 10 by 10 and
read 10 integers into a[0][0] to a[0][9].Then repeat the
insertion specified by an integer sequence nine times,and
store the results from a[1] to a[9].Finally,print the array.
Output Example:
1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3)
-------------------------------------------------------
I did as :
#include <stdio.h>
int main()
{
int a[10][10];
int i,j,k;
int x;
for(i=0; i<10; i++){
for(j=0; j<10; j++) a[i][j]=j+1;
}
for(i=0; i<10; i++){
for(j=0; j<i+1; j++){
x= a[i][j];
a[i][j] = a[i][j];
a[i][j]=x;
for(k=0; k<10; k++)printf("%d", a[j][k]);
printf("\n");
}
printf("\n");
}
return 0;
}
but,the output is compeletly different.
tanx in advance
推荐答案
" engartte" < NA ***** @ hotmail.com>在消息中写道
news：11 ********************** @ c13g2000cwb.googlegr oups.com ...
"engartte" <na*****@hotmail.com> wrote in message
news:11**********************@c13g2000cwb.googlegr oups.com...
大家好，
我在下面的问题上做了解决方案，但没有想要的输出。如果你知道某些方法，请告诉我它的评论。
- -------------------------------------------------- -
（声明一个二维数组a，大小为10×10，将10个整数读成[0] [0]到[0] [9]。然后重复
由整数序列指定9次插入，并将结果从[1]存储到[9]。最后，打印数组。
输出示例：
1 2 3 4 5 6 7 8 9 10
2
4
5
7
8
9 /> 10
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3）
---------------------------------------------- ---------
我做了：
#include< stdio.h>
int main（）
{
int a [10] [10];
int i，j，k;
int x;
for（i = 0; I< 10; i ++）{
for（j = 0; j< 10; j ++）a [i] [j] = j + 1;
}
i = 0; i< 10; i ++）{
for（j = 0; j< i + 1; j ++）{
x = a [i] [j];
a [ i] [j] = a [i] [j];
a [i] [j] = x;
for（k = 0; k <10; k ++）printf（"％d" ;，[j] [k]）;
printf（" \ n"）;
}
printf（" \ n"）;
}
返回0;
}
但是，输出完全不同。
tanx提前
hi all,
I did the solution on the following question,but no desired output is
acheived.if you know some way on,please show me with its comments.
-----------------------------------------------------
(Declare a two-dimensional array"a" of size 10 by 10 and
read 10 integers into a[0][0] to a[0][9].Then repeat the
insertion specified by an integer sequence nine times,and
store the results from a[1] to a[9].Finally,print the array.
Output Example:
1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3)
-------------------------------------------------------
I did as :
#include <stdio.h>
int main()
{
int a[10][10];
int i,j,k;
int x;
for(i=0; i<10; i++){
for(j=0; j<10; j++) a[i][j]=j+1;
}
for(i=0; i<10; i++){
for(j=0; j<i+1; j++){
x= a[i][j];
a[i][j] = a[i][j];
a[i][j]=x;
for(k=0; k<10; k++)printf("%d", a[j][k]);
printf("\n");
}
printf("\n");
}
return 0;
}
but,the output is compeletly different.
tanx in advance
你真的想用二维阵列存储什么？
你能说清楚吗，它是一个预定义的模式/序列吗？
-Neo
What you actually want to store in 2-D array?
Would you make it clear, is it a predefined pattern/sequence?
-Neo
engartte写道：
engartte wrote:
我在下面的问题上做了解决方案，但没有想要的输出是
实现的。如果你知道某种方式，请告诉我它的评论。
-------------------------------------- ---------------
（声明一个二维数组a，大小为10乘以10，并将10个整数读成[0] [0 ]到[0] [9]。然后重复由整数序列指定的
插入九次，并将结果从[1]存储到[9]。最后，打印数组。
输出示例：
1 2 3 4 5 6 7 8 9 10
2
3
5
6 > 7
8
9
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3）
-------------------------------------- -----------------
I did the solution on the following question,but no desired output is
acheived.if you know some way on,please show me with its comments.
-----------------------------------------------------
(Declare a two-dimensional array"a" of size 10 by 10 and
read 10 integers into a[0][0] to a[0][9].Then repeat the
insertion specified by an integer sequence nine times,and
store the results from a[1] to a[9].Finally,print the array.
Output Example:
1 2 3 4 5 6 7 8 9 10
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
2 1 3 4 5 6 7 8 9 10
1 3 2 4 5 6 7 8 9 10
3 2 4 1 5 6 7 8 9 10
2 4 1 5 3 6 7 8 9 10
4 1 5 3 6 2 7 8 9 10
1 5 3 6 2 7 4 8 9 10
5 3 6 2 7 4 8 1 9 10
3 6 2 7 4 8 1 9 5 10
6 2 7 4 8 1 9 5 10 3)
-------------------------------------------------------
#include< stdio.h>
void println（int a [10]）
{
int i = 10;
do printf（"％d" ，* a ++）; while（ - i&& printf（""））;
printf（" \ n"）;
}
int main（）
{
int a [10] [10] = {1,2,3,4,5,6， 7,8,9,10};
int i，j，k;
println（a [0]）;
for（i = 0; ++ i< 10;）
{
printf（"％d \ n"，k = a [0] [i]）;
--k;
for（j = 0; j< k; ++ j）a [i] [j] = a [ i-1] [j + 1];
a [i] [j] = a [i-1] [0];
while（++ j< 10）a [i] [j] = a [i-1] [j];
}
for（i = 0; i< 10; ++ i）println（a [i]）;
返回0;
}
/ *随意添加更多评论对此。 * /
#include <stdio.h>
void println(int a[10])
{
int i = 10;
do printf("%d", *a++); while (--i && printf(" "));
printf("\n");
}
int main()
{
int a[10][10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int i, j, k;
println(a[0]);
for (i = 0; ++i < 10; )
{
printf("%d\n", k = a[0][i]);
--k;
for (j = 0; j < k; ++j) a[i][j] = a[i-1][j+1];
a[i][j] = a[i-1][0];
while (++j < 10) a[i][j] = a[i-1][j];
}
for (i = 0; i < 10; ++i) println(a[i]);
return 0;
}
/* Feel free to add more comments to this. */
亲爱的Dietmar Schindler，
非常感谢解决方案。但是，如果可能的话，请告诉我更多关于
println
以及此程序的操作。对我来说，使用
术语println是新的因为我是C和阵列的新手。
提前谢谢
engartte
Dear Dietmar Schindler,
Thanks a lot for solution.but,If possible,please tell me more about
"println"
and the operation of this program.as for me, the use of the
term"println" is
new because I am new with C and array.
Thanks in advance
engartte
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