题目:有1、2、3、4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?

2025-04-08 07:18:26
推荐回答(4个)
回答1:

按数学算法就是 1X2X3X4 =24个不重复的 3位数

注意去掉for 括号后面的分号

1,2,3
1,2,4
1,3,2
1,3,4
1,4,2
1,4,3
2,1,3
2,1,4
2,3,1
2,3,4
2,4,1
2,4,3
3,1,2
3,1,4
3,2,1
3,2,4
3,4,1
3,4,2
4,1,2
4,1,3
4,2,1
4,2,3
4,3,1
4,3,2
Press any key to continue

#include
main()
{
int i,j,k;
for(i=1;i<5;i++)
for(j=1;j<5;j++)
for(k=1;k<5;k++)
{
if(i!=j&&j!=k&&k!=i)
{
printf("%d,%d,%d\n",i,j,k);
}
}
}

回答2:

123 124 132 134 142 143
213 214 231 234 241 243
312 314 321 324 341 342
412 413 421 423 431 432 共24个

回答3:

我把你的程序改了一下,你去运行一下就有结果了。 你的程序for循环后面不应该有分号
main()
{
int i,j,k,sum=0;
for(i=1;i<5;i++)
for(j=1;j<5;j++)
for(k=1;k<5;k++)
{
if(i!=j&&j!=k&&k!=i)
{
sum=i*100+j*10+k;
printf("%d\n",sum);
}
}
}

回答4:

#include "stdio.h"
void main()
{
int i,j,k;
for(i=1;i<5;i++)//去掉分号
for(j=1;j<5;j++)//去掉分号
for(k=1;k<5;k++)//去掉分号
{
if(i!=j&&j!=k&&k!=i)
{
printf("%d%d%d ",i,j,k);
}
}
}

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