C语言回顾day5（第6章三种循环语句）

待我称王封你为后i 2023-07-03 06:15 35阅读 0赞

### 文章目录 ###

*  while循环（入口条件循环）
 *   *  scanf的返回值作为while条件
     *  空语句作为循环体：跳过整数输入 （这个功能也可以通过continue实现）
     *  浮点数的比较 fabs函数
     *  猜数字游戏
     *  死循环 while(1)
     *   *  到底什么是true (答案：非零就是true，0就是false)
         *  while(x) VS while(x!=0)
         *  优先级
 *  for循环（入口条件循环，too）
 *   *  示例：打印立方表
     *  示例：打印ASCII码
     *  示例：猜数字（printf作为for循环的表达式1 ）
     *  for的死循环
     *  逗号运算符在for循环的使用
     *  示例 数列求和（飞靶悖论）
     *  for循环使用数组
     *  for循环写幂函数
 *  do while循环（出口条件循环）
 *   *  示例 猜数字（和上面的while实现对比要简短很多）
 *  如何选择循环
 *  关系运算符
 *  嵌套循环（nested loop）
 *  数组（循环经常使用数组）
 *  递增递减运算符进一步区分的示例
 *  练习

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70]

# while循环（入口条件循环） #

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 1]  
![在这里插入图片描述][20200203133302909.png]![在这里插入图片描述][20200203134244521.png]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 2]

## scanf的返回值作为while条件 ##

这个求和程序非常人性化，可以加任意数量的数，并且可以通过按键q退出，status变量是scanf的返回值，当输入的是数字，scanf就会获得一个值，于是返回值是1，而输入任何字母，在这个程序中不一定是q哦，就会使得status不是1，于是退出了while循环

#include <stdio.h>
    int main()
    { 
        int num;
        int sum = 0;
        int status;
        printf("Enter an integer to be summed:(q to quit)\n");
        status = scanf("%d", &num);
        while(status==1)
        { 
            sum += num;
            printf("enter next integer (q to quit):\n");
            status = scanf("%d", &num);
        }
        printf("These integers sum to %d.\n", sum);
        return 0;
    }

Enter an integer to be summed:(q to quit)
    12
    enter next integer (q to quit):
    23
    enter next integer (q to quit):
    q
    These integers sum to 35.

Enter an integer to be summed:(q to quit)
    12
    enter next integer (q to quit):
    23
    enter next integer (q to quit):
    w
    These integers sum to 35.

![在这里插入图片描述][20200203133721132.png]

上面的程序可以进一步简化，不必创建status变量

#include <stdio.h>
    int main()
    { 
        int num;
        int sum = 0;
        printf("Enter an integer to be summed:(q to quit)\n");
        while(scanf("%d", &num)==1)
        { 
            sum += num;
            printf("enter next integer (q to quit):\n");
        }
        printf("These integers sum to %d.\n", sum);
        return 0;
    }

![在这里插入图片描述][20200203133947192.png]  
如果把==写为=，就会一直让你输入

#include <stdio.h>
    int main()
    { 
        int num;
        int sum = 0;
        int status;
        printf("Enter an integer to be summed:(q to quit)\n");
        status = scanf("%d", &num);
        while(status=1)
        { 
            sum += num;
            printf("enter next integer (q to quit):\n");
            status = scanf("%d", &num);
        }
        printf("These integers sum to %d.\n", sum);
        return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 3]

## 空语句作为循环体：跳过整数输入 （这个功能也可以通过continue实现） ##

#include <stdio.h>
    int main()
    { 
        int num;
        printf("Enter an integer:\n");
        while(scanf("%d", &num)==1)
            ;
        return 0;
    }

只要输入的是数字，就不会终止while，而输入除了数字以外的字符，比如逗号，字母，百分号等等等等，就会终止while

Enter an integer:
    129854637w

![在这里插入图片描述][20200203135030618.png]  
关系运算符用于构建关系表达式  
![在这里插入图片描述][2020020313541379.png]![在这里插入图片描述][20200203135422137.png]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 4]

## 浮点数的比较 fabs函数 ##

#include <stdio.h>
    #include <math.h>
    int main()
    { 
        const double PI = 3.141592657;
        double response;
        printf("what is the value of pi?\n");
        scanf("%lf", &response);
        while(fabs(response - PI)> 0.001){ 
                printf("Try again!\n");
                scanf("%lf", &response);
        }
        printf("Close enough！\n");
        return 0;
    }

what is the value of pi?
    3.14
    Try again!
    3.141
    Close enough！

## 猜数字游戏 ##

#include <stdio.h>
    #include <math.h>
    int main()
    { 
        int number;
        printf("enter a number:\n");
        scanf("%d", &number);
        while(number < 10)
        { 
            printf("your number is too small!\n");
            scanf("%d", &number);
        }
        printf("now it's ok.\n");
        return 0;
    }

enter a number:
    2
    your number is too small!
    5
    your number is too small!
    7
    your number is too small!
    11
    now it's ok.

## 死循环 while(1) ##

![在这里插入图片描述][20200203141433389.png]  
![在这里插入图片描述][20200203141513774.png]

### 到底什么是true (答案：非零就是true，0就是false) ###

#include <stdio.h>
    int main()
    { 
        int n = 3;
        while(n)
            printf("%d is true.\n", n--);
    	printf("%d is false.\n", n);
        n = -3;
        while(n)
            printf("%d is true.\n", n++);
        printf("%d is false.\n", n);
        return 0;
    }

3 is true.
    2 is true.
    1 is true.
    0 is false.
    -3 is true.
    -2 is true.
    -1 is true.
    0 is false.

### while(x) VS while(x!=0) ###

他们的作用是完全一样的

初学者喜欢用第二种，容易理解

但是程序员更喜欢第一种，简洁  
![在这里插入图片描述][20200203142404882.png]

### 优先级 ###

![在这里插入图片描述][20200203153035854.png]

x>y+2 相当于 x>(y+2)  
![在这里插入图片描述][20200203153236966.png]

# for循环（入口条件循环，too） #

![在这里插入图片描述][20200203154205726.png]

下面这些示例说明了for循环的灵活性，三个表达式修修改改，就可以很灵活

## 示例：打印立方表 ##

#include <stdio.h>
    int main()
    { 
        printf("i i cubed\n");
        for(int i=0;i<6;i++)
            printf("%-5d %-5d\n", i, i*i*i);
        return 0;
    }

i      i cubed
    0      0
    1      1
    2      8
    3      27
    4      64
    5      125

## 示例：打印ASCII码 ##

#include <stdio.h>
    int main()
    { 
        char ch;
        for(ch='a';ch<='z';ch++)
            printf("The ASCII code for %c is %d.\n", ch, ch);
        return 0;
    }

The ASCII code for a is 97.
    The ASCII code for b is 98.
    The ASCII code for c is 99.
    The ASCII code for d is 100.
    The ASCII code for e is 101.
    The ASCII code for f is 102.
    The ASCII code for g is 103.
    The ASCII code for h is 104.
    The ASCII code for i is 105.
    The ASCII code for j is 106.
    The ASCII code for k is 107.
    The ASCII code for l is 108.
    The ASCII code for m is 109.
    The ASCII code for n is 110.
    The ASCII code for o is 111.
    The ASCII code for p is 112.
    The ASCII code for q is 113.
    The ASCII code for r is 114.
    The ASCII code for s is 115.
    The ASCII code for t is 116.
    The ASCII code for u is 117.
    The ASCII code for v is 118.
    The ASCII code for w is 119.
    The ASCII code for x is 120.
    The ASCII code for y is 121.
    The ASCII code for z is 122.

## 示例：猜数字（printf作为for循环的表达式1 ） ##

这个应用很容易想到用while去做，不过for也可以做到，第一个表达式在每次循环被执行一次。  
![在这里插入图片描述][20200203155820665.png]

#include <stdio.h>
    int main()
    { 
        int num;
        for(printf("Keep entering numbers!\n");num!=10;)
            scanf("%d", &num);
        printf("That's the number I want!\n");
        return 0;
    }

Keep entering numbers!
    1
    2
    6
    9
    4
    10
    That's the number I want!

## for的死循环 ##

![在这里插入图片描述][2020020316005619.png]

#include <stdio.h>
    int main()
    { 
        char ch;
        for(;;)
            printf("I won't stop!\n");
        return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 5]

## 逗号运算符在for循环的使用 ##

![在这里插入图片描述][20200203172358241.png]

关于逗号运算符的使用，最重要的是它有个顺序限制，限制自己左边的表达式先执行，就这么一点  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 6]![在这里插入图片描述][20200203173023980.png]  
但是要注意判断条件（表达式2）还是只能写一个的哈，初始值（表达式1）和操作（表达式3）可以通过逗号运算符写多个

#include <stdio.h>
    int main()
    { 
        printf("x y\n");
        for(int x = 1, y = 10; x < 10; x++, y--)
            printf("%d %d\n", x, y);
        return 0;
    }

x  y
    1  10
    2  9
    3  8
    4  7
    5  6
    6  5
    7  4
    8  3
    9  2

![在这里插入图片描述][20200203172138556.png]

## 示例 数列求和（飞靶悖论） ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 7]  
![在这里插入图片描述][20200203173846577.png]

#include <stdio.h>
    int main()
    { 
        double sum = 0.0;
        printf("i sum\n");
        for(int i =1; i<500; i*=2)
        { 
            sum += 1.0/i;
            printf("%-2d %-4.3f\n", i, sum);
        }
        return 0;
    }

可以看到，结果很接近2，数学里这个无穷数列的和是2. 所以这个飞靶在有限时间里一定可以射到目的地·····

飞靶悖论破解····

i     sum
    1   1.000
    2   1.500
    4   1.750
    8   1.875
    16  1.938
    32  1.969
    64  1.984
    128  1.992
    256  1.996

## for循环使用数组 ##

注意的是scanf输入给数组的单个元素时还是需要&运算符的  
转换说明要和数据类型对应起来

/*使用循环来处理数组*/
    #include <stdio.h>
    #define PAR 72
    #define SIZE 10
    int main()
    { 
        int index, score[SIZE], sum = 0;
        float average, handicap;
        printf("Enter %d golf scores:\n", SIZE);
        for(index=0;index<SIZE;index++)
            scanf("%d", &score[index]);
        printf("The input scores are:\n");
        for(index=0;index<SIZE;index++)
        { 
            printf("%5d", score[index]);
            sum += score[index];
        }
        printf("\n");
        printf("The sum of the %d scores is %d.\n", SIZE, sum);
        average = (float)sum/SIZE; //
        handicap = PAR - average;
        printf("The average is %4.2f, and the handicap is %4.2f.\n", average, handicap);
        return 0;
    }

Enter 10 golf scores:
    1 2 3 4 5 6 7 8 9 10
    The input scores are:
        1    2    3    4    5    6    7    8    9   10
    The sum of the 10 scores is 55.
    The average is 5.50, and the handicap is 66.50.

## for循环写幂函数 ##

这个例子我翻了一次错，以前也犯过的，就是scanf的两个输入变量之间不能有逗号，如果用逗号隔开那么你自己输入数据的时候必须也用逗号隔开！！！如果你用空格在scanf的格式字符串中分隔数据那么在程序运行时也用空格隔开，必须保持一致！！！！  
一般我会习惯于在写程序时在scanf用逗号分隔数据，却在程序运行时又用空格分隔！！于是就造成了错误！！这一点很重要，务必记住！

/*幂函数的实现*/
    #include <stdio.h>
    double pow(double p, int n); // 原型，p的n次方
    int main()
    { 
        double x;
        int y;
        printf("enter a number and a positive integer power:\n");
        scanf("%lf %d", &x, &y);
        printf("The result is %4.2f", pow(x, y));
        return 0;
    }
    double pow(double p, int n)
    { 
        double pow = 1.0;
        for(int i = 0;i<n;i++)
            pow *= p;
        return pow;
    }

![在这里插入图片描述][20200204132356700.png]

enter a number and a positive integer power:
    2.3 3
    The result is 12.17

下面是 scanf("%lf, %d", &x, &y);但是程序运行用空格分开得到的错误结果

enter a number and a positive integer power:
    2.3 3
    The result is 1.00

但是运行程序也用逗号就没错了

enter a number and a positive integer power:
    2.3,3
    The result is 12.17

# do while循环（出口条件循环） #

do while的关键就是至少执行一次循环体

注意while后面有分号  
![在这里插入图片描述][20200203174602143.png]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 8]

## 示例 猜数字（和上面的while实现对比要简短很多） ##

#include <stdio.h>
    int main()
    { 
        int num;
        do
        { 
            scanf("%d", &num);
        }while(num!=10);
        printf("That's the number I want!");
        return 0;
    }

1
    2
    3
    6
    10

# 如何选择循环 #

![在这里插入图片描述][20200203175048694.png]  
![在这里插入图片描述][20200203175124148.png]  
需要必须至少执行一次循环体的场景还是很少的，输出提示语是一个例子

那么入口循环的for 和 while怎么选择呢？

![在这里插入图片描述][20200203175221247.png]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 9]

# 关系运算符 #

![在这里插入图片描述][20200203173113251.png]

这里的组合运算符说的是+= -= /= \*= %=一类的

# 嵌套循环（nested loop） #

![在这里插入图片描述][20200203180459681.png]![在这里插入图片描述][20200203180735839.png]![在这里插入图片描述][20200203180741471.png]  
示例

#include <stdio.h>
    int main()
    { 
        for(int row = 1; row < 5; row++)
        { 
            for(char ch = 'A'; ch < (10 + 'A'); ch++)
                printf("%c", ch);
            printf("\n");
        }
    
        return 0;
    }

ABCDEFGHIJ
    ABCDEFGHIJ
    ABCDEFGHIJ
    ABCDEFGHIJ

稍微改动一点点

#include <stdio.h>
    int main()
    { 
        for(int row = 1; row < 10; row++)
        { 
            for(char ch = 'A' + row; ch < (10 + 'A'); ch++)
                printf("%c", ch);
            printf("\n");
        }
        return 0;
    }

BCDEFGHIJ
    CDEFGHIJ
    DEFGHIJ
    EFGHIJ
    FGHIJ
    GHIJ
    HIJ
    IJ
    J

# 数组（循环经常使用数组） #

![在这里插入图片描述][20200203183746401.png]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 10]![在这里插入图片描述][20200203183911675.png]

# 递增递减运算符进一步区分的示例 #

#include <stdio.h>
    #define FORMAT "%s! C is cool!\n"
    int main()
    { 
        int num = 10;
        printf(FORMAT, FORMAT);
        printf("%d\n", ++num);
        printf("%d\n", num++);
        printf("%d\n", --num);
        printf("%d\n", num--);
        return 0;
    }

%s! C is cool!
    ! C is cool!
    11
    11
    11
    11

# 练习 #

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 11]

#include <stdio.h>
    int main()
    { 
        for(int i=0;i<4;i++)
        { 
            for(int j=0;j<8;j++)
                printf("%c", '$');
            printf("\n");
        }
        return 0;
    }

$$$$$$$$
    $$$$$$$$
    $$$$$$$$
    $$$$$$$$

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 12]

#include <stdio.h>
    int main()
    { 
        for(int i=0;i<6;i++)
        { 
            for(char ch = 'F';ch>='F'-i;ch--)
                printf("%c", ch);
            printf("\n");
        }
        return 0;
    }

F
    FE
    FED
    FEDC
    FEDCB
    FEDCBA

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 13]

#include <stdio.h>
    int main()
    { 
         char ch = 'A';
         char ch1;
        for(int i=0;i<6;i++)
        { 
            ch = ch + i;
            for(ch1=ch;ch1<=ch+i;ch1++)
                printf("%c", ch1);
            printf("\n");
        }
        return 0;
    }

A
    BC
    DEF
    GHIJ
    KLMNO
    PQRSTU

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 14]

#include <stdio.h>
    int main()
    { 
        char ch;
        printf("Enter a capital letter:\n");
        scanf("%c", &ch);
        for(int i=0;i<=ch-'A';i++)
        { 
            //打印空格
            for(int j=0;j<ch-'A'-i;j++)
                printf(" ");
            //升序字母
            for(int j=0;j<=i;j++)
                printf("%c", 'A'+j);
            //降序字母
            for(int j=0;j<i;j++)
                printf("%c", 'A'+i-j-1);
            //空格
            for(int j=0;j<ch-'A'-i;j++)
                printf(" ");
            printf("\n");
        }
        return 0;
    }

Enter a capital letter:
    E
        A
       ABA
      ABCBA
     ABCDCBA
    ABCDEDCBA

Enter a capital letter:
    W
                          A
                         ABA
                        ABCBA
                       ABCDCBA
                      ABCDEDCBA
                     ABCDEFEDCBA
                    ABCDEFGFEDCBA
                   ABCDEFGHGFEDCBA
                  ABCDEFGHIHGFEDCBA
                 ABCDEFGHIJIHGFEDCBA
                ABCDEFGHIJKJIHGFEDCBA
               ABCDEFGHIJKLKJIHGFEDCBA
              ABCDEFGHIJKLMLKJIHGFEDCBA
             ABCDEFGHIJKLMNMLKJIHGFEDCBA
            ABCDEFGHIJKLMNONMLKJIHGFEDCBA
           ABCDEFGHIJKLMNOPONMLKJIHGFEDCBA
          ABCDEFGHIJKLMNOPQPONMLKJIHGFEDCBA
         ABCDEFGHIJKLMNOPQRQPONMLKJIHGFEDCBA
        ABCDEFGHIJKLMNOPQRSRQPONMLKJIHGFEDCBA
       ABCDEFGHIJKLMNOPQRSTSRQPONMLKJIHGFEDCBA
      ABCDEFGHIJKLMNOPQRSTUTSRQPONMLKJIHGFEDCBA
     ABCDEFGHIJKLMNOPQRSTUVUTSRQPONMLKJIHGFEDCBA
    ABCDEFGHIJKLMNOPQRSTUVWVUTSRQPONMLKJIHGFEDCBA

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 15]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 16]![在这里插入图片描述][20200204133430756.png]

#include <stdio.h>
    #include <math.h>
    int main()
    { 
        int x[8];
        for(int i=0;i<8;i++)
            x[i] = pow(2,i);
        int j=0;
        do{ 
            printf("%d ", x[j++]);
        }while(j<8);
        return 0;
    }

1 2 4 8 16 32 64 128

![在这里插入图片描述][20200204133234198.png]

#include <stdio.h>
    int main()
    { 
        printf("year money\n");
        for(int year = 1, money = 1000000;money>0;year++)
        { 
            money = money*1.08-100000;
            printf("%-5d%-3d\n", year, money);
            if(money<=0)
                printf("After %d years, The account was drained of money.\n", year);
        }
        return 0;
    }

year money
    1    980000
    2    958400
    3    935072
    4    909877
    5    882667
    6    853280
    7    821542
    8    787265
    9    750246
    10   710265
    11   667086
    12   620452
    13   570088
    14   515695
    15   456950
    16   393506
    17   324986
    18   250984
    19   171062
    20   84746
    21   -8474
    After 21 years, The account was drained of money.

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 17]  
18题和17题一毛一样

#include <stdio.h>
    int main()
    { 
        printf("week friends\n");
        for(int week = 1, friends = 5;friends<150;week++)
        { 
            friends = (friends - week) * 2;
            printf("%-5d%-3d\n", week, friends);
        }
        return 0;
    }

week friends
    1    8
    2    12
    3    18
    4    28
    5    46
    6    80
    7    146
    8    276

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70]: https://img-blog.csdnimg.cn/20200203123900303.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20200203154139385.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203133302909.png]: https://img-blog.csdnimg.cn/20200203133302909.png
[20200203134244521.png]: https://img-blog.csdnimg.cn/20200203134244521.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 2]: https://img-blog.csdnimg.cn/20200203133326226.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203133721132.png]: https://img-blog.csdnimg.cn/20200203133721132.png
[20200203133947192.png]: https://img-blog.csdnimg.cn/20200203133947192.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 3]: https://img-blog.csdnimg.cn/20200203152414729.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203135030618.png]: https://img-blog.csdnimg.cn/20200203135030618.png
[2020020313541379.png]: https://img-blog.csdnimg.cn/2020020313541379.png
[20200203135422137.png]: https://img-blog.csdnimg.cn/20200203135422137.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 4]: https://img-blog.csdnimg.cn/20200203135510977.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203141433389.png]: https://img-blog.csdnimg.cn/20200203141433389.png
[20200203141513774.png]: https://img-blog.csdnimg.cn/20200203141513774.png
[20200203142404882.png]: https://img-blog.csdnimg.cn/20200203142404882.png
[20200203153035854.png]: https://img-blog.csdnimg.cn/20200203153035854.png
[20200203153236966.png]: https://img-blog.csdnimg.cn/20200203153236966.png
[20200203154205726.png]: https://img-blog.csdnimg.cn/20200203154205726.png
[20200203155820665.png]: https://img-blog.csdnimg.cn/20200203155820665.png
[2020020316005619.png]: https://img-blog.csdnimg.cn/2020020316005619.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 5]: https://img-blog.csdnimg.cn/20200203160143315.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203172358241.png]: https://img-blog.csdnimg.cn/20200203172358241.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 6]: https://img-blog.csdnimg.cn/20200203172328508.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203173023980.png]: https://img-blog.csdnimg.cn/20200203173023980.png
[20200203172138556.png]: https://img-blog.csdnimg.cn/20200203172138556.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 7]: https://img-blog.csdnimg.cn/20200203173903986.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203173846577.png]: https://img-blog.csdnimg.cn/20200203173846577.png
[20200204132356700.png]: https://img-blog.csdnimg.cn/20200204132356700.png
[20200203174602143.png]: https://img-blog.csdnimg.cn/20200203174602143.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 8]: https://img-blog.csdnimg.cn/2020020317461973.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203175048694.png]: https://img-blog.csdnimg.cn/20200203175048694.png
[20200203175124148.png]: https://img-blog.csdnimg.cn/20200203175124148.png
[20200203175221247.png]: https://img-blog.csdnimg.cn/20200203175221247.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 9]: https://img-blog.csdnimg.cn/20200203175313110.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203173113251.png]: https://img-blog.csdnimg.cn/20200203173113251.png
[20200203180459681.png]: https://img-blog.csdnimg.cn/20200203180459681.png
[20200203180735839.png]: https://img-blog.csdnimg.cn/20200203180735839.png
[20200203180741471.png]: https://img-blog.csdnimg.cn/20200203180741471.png
[20200203183746401.png]: https://img-blog.csdnimg.cn/20200203183746401.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 10]: https://img-blog.csdnimg.cn/20200203183856219.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200203183911675.png]: https://img-blog.csdnimg.cn/20200203183911675.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 11]: https://img-blog.csdnimg.cn/20200204132600416.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 12]: https://img-blog.csdnimg.cn/20200204133107288.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 13]: https://img-blog.csdnimg.cn/20200204133043680.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 14]: https://img-blog.csdnimg.cn/20200204133023391.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 15]: https://img-blog.csdnimg.cn/20200204133547729.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 16]: https://img-blog.csdnimg.cn/20200204133451685.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70
[20200204133430756.png]: https://img-blog.csdnimg.cn/20200204133430756.png
[20200204133234198.png]: https://img-blog.csdnimg.cn/20200204133234198.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 17]: https://img-blog.csdnimg.cn/2020020413322265.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0,size_16,color_FFFFFF,t_70