（for述句）韩信点兵中的最大值问题

#include <stdio.h>
int main()
{


int max;
printf("MAX = ");
scanf("%d", &max);
int number;
int answer = 0;
for (number = max;number >= 1 && answer == 0; --number)
{
    if (number%3 == 2 && number%5 == 3 && number%7 == 2)
    {
        answer = number;
    }
}
if (answer != 0)
{
  printf("%d\n",answer) ;
}

return 0;
}

 韩信点兵找第三个小的练习

#include <stdio.h>
int main()
{
    int count = 0;
    int number;
    for (number = 1; number <= 1000 && count <3;++number)
    {
        if (number%3 == 2 && number%5 == 3 && number%7 ==2)
        {
            ++count;
            if (count == 3)
            {
                printf("%d", number);
            }
        }
    }
    return 0;
}

 10.6质数判断练习

1.

#include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);
    int number;
    for (number = 2; number < N && N % number != 0; ++number);

    if (number == N)
    {
        printf("Yes\n");
    }else
    {
        printf("No\n");
    }
    return 0;

}

N = 7
Yes

Process returned 0 (0x0)   execution time : 2.835 s
Press any key to continue.

2.

#include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);

    int isPrime = 1;
    int number;
    for (number = 2; number < N && isPrime; ++number)//这里的&& isPrime 可以省略
    {
        if (N % number == 0)
    {
        isPrime = 0;
    }
    }
    if (isPrime) {
        printf("Yes\n");
    } else {
    printf("No\n");
    }
    return 0;
}

 11.1 9x9连乘（9X9的正方形）

#include <stdio.h>
int main()
{
    int i,j;
    for (i =2; i <= 9; ++i)
    {
        for (j = 1; j <= 9; ++j)
        {
            printf("%d x %d = %d\n",i,j, i*j); // printf("*");
        }// printf("\n");
    }
    return 0;
}

11.2 用文字绘制空心长方形的练习

# include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);

    int i,j;
    for (i = 1; i <= N; ++i)
    {
        for (j = 1; j <= N; ++j)
        {
            if (i == 1 || i == N || j == 1 || j == N) // 这一步操作使它空心
            {
                printf("*");

            } else {
                printf(" ");
            }
        }
        printf("\n");

    }
    return 0;
}

N = 7
*******
*     *
*     *
*     *
*     *
*     *
*******

Process returned 0 (0x0)   execution time : 3.764 s
Press any key to continue.

三角形

# include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);

    int i,j;
    for (i = 1; i <= N; ++i)
    {
        for (j = 1; j <= i; ++j)
        {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

N = 7
*
**
***
****
*****
******
*******

Process returned 0 (0x0)   execution time : 3.901 s
Press any key to continue.

11.4 用文字绘制三角形的练习

# include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);

    int i,j;
    for (i = 1; i <= N; ++i)
    {
        for (j = 1; j <= i; ++j)
        {
            printf("*");
        }
        printf("\n");
    }
    return 0;
}

印 1 个星星                                  + 换行
印 1 个星星 + i个空白 + 1 个星星 + 换行 (i=0)
印 1 个星星 + i个空白 + 1 个星星 + 换行 (i=1)
印 1 个星星 + i个空白 + 1 个星星 + 换行 (i=2)
印 1 个星星 + i个空白 + 1 个星星 + 换行 

 

11.5 基于坐标法用文字绘制三角形

#include <stdio.h>
int main()
{
    int N;
    printf("N = ");
    scanf("%d", &N);

    int i,j;
    for (i = 1; i <= N; ++i)
    {
        for (j = 1; j <= N; ++j)
        if (i+j >= N+1)
        {
            printf("*");

        }else {
            printf(" ");

        }
        printf("\n");
    }

    return 0;
}
      *
     **
    ***
   ****
  *****
 ******
*******
// 可以有多种变化，只要看i，j变化

 11.6 找简易方程式整数解（两种方法）

#include <stdio.h>
int main()
{
    int i,j;
    for (i = 1; i <= 30; ++i)
    {
        for (j =1; j <= 30; ++j)
            if (i+j == 30 && i*j == 221)
        {
            printf("%d, %d\n",i,j);
        }
    }
    return 0;
}


#include <stdio.h>
int main()
{
    int i;
    for (i=0;i <= 30/2; ++i)
    {
        int j = 30 - i;
        if (i*j == 221)
        {
            printf("%d, %d\n", i,j);
        }

    }

    return 0;
}