编译代码，输出乘法口诀
#include <stdio.h>
#include <conio.h>
int main()
{
    int i=1,j=1;
    for(i=1;i<=9;i++)
    {

        if(j==1)
        {

           for(j=1;j<2;j++)
              {

            int res;
            res=i*j;
            printf("%d*%d=%-3d ",j,i,res);
        }

        printf("\n");
        }

        else if(j>=1 && j<=2)
        {
            for(j=1;j<3;j++)
            {

                int res;
                res=i*j;
            printf("%d*%d=%-3d ",j,i,res);
        }

        printf("\n");
        }
        else if(j>=1 && j<=3)
        {
            for(j=1;j<4;j++)
                {
                    int res;
                    res=i*j;
                    printf("%d*%d=%-3d ",j,i,res);
                }
                printf("\n");
            }
        else if(j>=1 && j<=4)
        {
            for(j=1;j<5;j++)
                {
                    int res;
                    res=i*j;
                    printf("%d*%d=%-3d ",j,i,res);
                }
                printf("\n");
            }

        else if(j>=1&&j<=5)
        {
            for(j=1;j<6;j++)
                {
                    int res;
                    res=i*j;
                    printf("%d*%d=%-3d",j,i,res);
                }
                printf("\n");
            }

        else if(j>=1 && j<=6)
        {
            for(j=1;j<7;j++)
                {
                    int res;
                    res=i*j;
                    printf("%d*%d=%-3d",j,i,res);
                }
                printf("\n");
            }

            else if(j>=1 && j<=7)
            {
                for(j=1;j<8;j++)
                    {
                        int res;
                        res=i*j;
                        printf("%d*%d=%-3d",j,i,res);
                    }
                    printf("\n");
                }

            else if(j>=1 && j<=8)
            {
                for(j=1;j<9;j++)
                    {
                        int res;
                        res=i*j;
                        printf("%d*%d=%-3d",j,i,res);
                    }
                    printf("\n");
                }

            else if(j>=1  && j<=9)
            {
                for(j=1;j<10;j++)
                    {
                        int res;
                        res=i*j;
                        printf("%d*%d=%-3d",j,i,res);
                    }
                    printf("\n");
                }



        }

        printf("\n");

    getch();
    return 0;
}

这种写法成什么话，用双重嵌套控制两个因数啊

代码可以优化
#include <stdio.h>
main()
{
    int i, j;
    for (i=1; i<10; ++i)
    {
        for (j=1; j<=i; ++j)
        {
            printf("%d*%d=%d\t", j,i,i*j);
        }
        printf("\n");
    }
}

#include "stdio.h"
void f(int i, int j){
    if(j>0){
        f(i, j-1);
        printf("%d*%d=%2d\t", i, j, i * j);
    }
}
void f(int i){
    if(i>0){
        f(i-1);
        f(i,i);
        printf("\n");
    }
}
int main(){
    f(9);
} 

感谢，看了你们写的，再看看我写的，真的可以用乱码来形容了。。。

勉強すればするほど、難しくなる感じがします。我慢強くて続けているよ、

发些人家看不懂的东西有意思么~不知道的人还以为是乱码~曾经有些贴因为乱码的原因被删了~~

好哒，下次改下输入法设置。不跟大神聊咯，追剧去了，。。

无聊闲逛来灌水~~
看出你应该是学java的吧~学java就学java~~怎么把java的格式搬上C了？~~那种写法C过不了~得要改个函数名~~

PS:

1:当然C++也支持这种写法~不过如果 int main末尾没有加return 0;则会有警告(C却无语地没有要求)~如果不是非得用C++实现的还是尽量避免这种写法好~

2:递归虽然形式精美~不过递归有个出栈过程~这里效率是没有直接用两个for循环高的~~不过平时凑合看看也好~~~

[此贴子已经被作者于2017-5-31 00:03编辑过]

好像弱类型语言才不支持诶，java也看点没那么深入，该背单词咯