初学C,请帮忙!

KLite

我的九九乘法表是这样的:
main(void)
{
int i,j;
for(i=1;i<=9;i++)
{
  for(j=i;j<=9;j++)
  {
   printf("%d X %d= %d\t",i,j,i*j);
   printf("\n");
  }
}
}
输出成这样:
1 X 1= 1
1 X 2= 2
1 X 3= 3
1 X 4= 4
1 X 5= 5
1 X 6= 6
1 X 7= 7
1 X 8= 8
1 X 9= 9
2 X 2= 4
2 X 3= 6
2 X 4= 8
2 X 5= 10
2 X 6= 12
2 X 7= 14
2 X 8= 16
2 X 9= 18
3 X 3= 9
3 X 4= 12
3 X 5= 15
3 X 6= 18
3 X 7= 21
3 X 8= 24
3 X 9= 27
4 X 4= 16
4 X 5= 20
4 X 6= 24
4 X 7= 28
4 X 8= 32
4 X 9= 36
5 X 5= 25
5 X 6= 30
5 X 7= 35
5 X 8= 40
5 X 9= 45
6 X 6= 36
6 X 7= 42
6 X 8= 48
6 X 9= 54
7 X 7= 49
7 X 8= 56
7 X 9= 63
8 X 8= 64
8 X 9= 72
9 X 9= 81

我要的格式是这样:
1×1=1
1×2=2 2×2=4
1×3=3 2×3=6 3×3=9
1×4=4 2×4=8 3×4=12 4×4=16
1×5=5 2×5=10 3×5=15 4×5=20 5×5=25
1×6=6 2×6=12 3×6=18 4×6=24 5×6=30 6×6=36
1×7=7 2×7=14 3×7=21 4×7=28 5×7=35 6×7=42 7×7=49
1×8=8 2×8=16 3×8=24 4×8=32 5×8=40 6×8=48 7×8=56 8×8=64
1×9=9 2×9=18 3×9=27 4×9=36 5×9=45 6×9=54 7×9=63 8×9=72 9×9=81

该怎样修改?我试过好几次,都不成功,只好求助.我知道这问题很简单,但我是初学者,没办法,麻烦你了~

souky

main(void)
{
int i,j;
for(i=1;i<=9;i++)
{
for(j=1;j<=i;j++)
{
printf("%d X %d= %d\t",j,i,i*j);
}
printf("\n");
}
}
我改了一下，这样应该行，我没试过，不行再找我

CNOOC

C 我就不知道, 但如果你用C++可以这么做.
[code:1]#include <iostream>

using namespace std;

int main()
{
        int i, j;

        for(j=1; j<=9; j++)
        {
                for(i=1; i<=j; i++)
                {
                        cout << i << 'X' << j << '=' << i*j << ' ';
                }
                cout << "\n";
        }

        cin.get();cin.get();
        return 0;

}[/code:1]

KLite

[quote:953f26553b="souky"]main(void)
{
int i,j;
for(i=1;i<=9;i++)
{
for(j=1;j<=i;j++)
{
printf("%d X %d= %d\t",j,i,i*j);
}
printf("\n");
}
}
我改了一下，这样应该行，我没试过，不行再找我[/quote]
可以,呵呵,想不到这么简单!

KLite

还有这样一道题不会:
编一程序,输出2的1~10次方的值.

souky

main(void)
{
int i,j;
for(i=1;i<=10;i++)
{
a=1
for(j=1;1<=i;j++)
{
a=a*2
}
printf(a);
printf("\n");
}
}

在basic里如下：
For I=1 to 10
  A=1
For t=1 to I
   A=A*2
Next t
Print A
Next I

我喜欢用basic，用c语言写的程序不超过50个，呵呵

CNOOC

[code:1]#include <iostream>

using namespace std;

int main()
{
        int i, j;
        j=2;

        for (i=1; i<=10; i++)
        {
                cout << '2' << '^' << i << '=' << j << endl;
                j=j*2;
        }

        cin.get();cin.get();

        return 0;
}[/code:1]

CNOOC

or

[code:1]#include <iostream>
#include <cmath>

using namespace std;

int main()
{
        int y;

        for (y=1; y<=10; y++)
        {
                cout << '2' << '^' << y << "= " << pow(2, y) << endl;
        }

        cin.get();cin.get();
        return 0;
}[/code:1]

Dragonfly

a question, using ansi c, how to print till 2^48. and can u give a fastest algorithm?
making it out is one thing, making it fast is another thing.

CNOOC

Who are you asking? I don't think it will be me.

Dragonfly

anybody can answer this question, hehe. i want to know a correct and fast solution.
u int can not store 2^48. will overflow.

CNOOC

Then I will use a double, but I couldn't find the faster solution. What's your way? How do you do it?

Dragonfly

but i need a integer result. then how to do? and what is the max double? i forget. i have no solution. i am quite busy on my work now.

CNOOC

Since the question is asking the power for base 2, then number we get will be integer anyway, even though we sign double to it. the range of double value is (2.2e-308 to 1.8e30, and the long double will be (3.4e-4932 to 1.1e+4932)

Beside using pow(), I can try loop or recursion, but that only makes it longer. Basicly I am a blind mind man, so I haven't found other possible way yet.

CNOOC

Probably we can ask someone about this on www.cppcn.net , www.chinaunix.net , or www.unix.com