C语言第4天循环，流程控制。

C语言第四天

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【1】if

if (表达式) 

{语句块1}                            

else

{语句块2}

 

 

//else只跟上一条if有关

 

 

【2】

switch (表达式)

{  

case 常量表达式1 : 语句块1；break;

   case 常量表达式2 : 语句块2；break;

   …

  case 常量表达式n ：语句块n；break;

  default:语句块n+1；

  }

 

 

【3】goto

进行跳转的语句，需要定义标识符

 

【4】while

 

while（表达式）

{

循环体；

}

 

 

【5】for

 

for(expression1； expression2； expression3)

{

     statements;

}

 

1--- expression1：赋值语句,对循环变量初始化。最先执行，且只执行一次

2--- expression2：控制循环条件,为真：执行循环体，为假：跳出for

3--- expression3：一般用于，改变循环变量。执行完循环体后再执行；

 

 

 

_F

__FE

___FED

____FEDC

_____FEDCB

______FEDCBA

 

 

 

大小写转换代码<wiz_tmp_tag id="wiz-table-range-border" contenteditable="false" style="display: none;">

 

 

 

 

1. # include <stdio.h>
2. int main(int argc,constchar*argv[])
3. {
4. #if 0
5. int a =10;
6. if(a>0)
7. printf("a>0");
8. elseif(a==0)
9. printf("a=0");
10. else
11. printf("a<0");
12. putchar(10);
13. return0;
14. #endif
15. puts("please input word");
16. int a =0;
19. a= getchar();
21. if('A'<=a&&a<='Z')
22. {
23. putchar(a+32);
24. }
25. elseif( a>='a'&&a<='z')
26. {
27. putchar (a-32);
28. }
29. else
30. puts("error");
31. }

奇数偶数判断求平均值

1. #include<stdio.h>
3. int main(int argc,constchar*argv[])
4. {
5. int i =0, count_ji =0, count_ou =0, num_ji =0, num_ou =0;
7. while(1)
8. {
9. if(scanf("%d",&i)!=1)
10. {
11. puts("error");
12. return-1;
13. }
15. if(i ==0)
16. {
17. puts("end");
18. break;
19. }
21. if(i %2)
22. {
23. count_ji++;
24. num_ji += i;
25. }
26. else
27. {
28. count_ou++;
29. num_ou += i;
30. }
31. }
34. if(count_ji !=0)
35. {
36. num_ji /= count_ji;
37. }
39. if(count_ou !=0)
40. {
41. num_ou /= count_ou;
43. }
45. printf("count_ji = %d, count_ou = %d\n",count_ji, count_ou);
46. printf("平均值_ji = %d, 平均值_ou = %d\n",num_ji, num_ou);
48. return0;
49. }

输出分钟转小时

1. int a =3;
2. int i=0,j=0;
3. float sum_ji=0;
4. float sum_ou=0;
5. while(1)
6. {
7. if( scanf("%d",&a)!=1)
8. {
9. return-1;
10. }
11. else
12. if(a==0)
13. {
14. puts("end");
15. break;
16. }
17. if(a%2==0)
18. {
19. i++;
20. sum_ou+=a;
21. }
22. else
23. {
24. j++;
25. sum_ji+=a;
26. }
27. }
29. printf("%d %d\n",i,j);
30. if(i!=0)
31. {
32. printf("%f\n",sum_ou);
33. }
34. if(j!=0)
35. {
36. printf("%f\n",sum_ji);
37. }
38. #endif
40. int a;
41. int xiaoshi=0,fen=0;
42. while(1)
43. {
44. if(scanf("%d",&a)!=1)
45. {
47. printf("try again!");
48. getchar();
49. continue;
50. }
51. else
52. {
53. xiaoshi=a/60;
54. fen= a%60;
55. printf("%02d:%02d\n",xiaoshi,fen);
56. }
60. }
62. return0;
63. }

乘法解决

 

1. # include <stdio.h>
2. int main(int argc,constchar*argv[])
3. {
5. /* int i=0;
6. int sum =0;
7. for( i=0;i<=10;i++)
8. {
9. sum+=i;
10. }
11. printf("%d\n",sum);
13. */
17. int a,b,i,j;
18. for(i=1;i<=9;i++)
19. {
20. for(j=1;j<=i;j++)
21. {
22. printf("%-2d*%-2d=%-2d ",j,i,i*j);
24. }
25. putchar(10);
26. }
28. return0;
29. }

水仙花数 abc=a的三次+b的三次+c的三次

1. int i ;
2. int bai,shi,ge;
3. for(i=100;i<=999;i++)
4. {
5. ge = i%10;
6. shi=i/10%10;
7. bai=i/100;
8. // printf("%d\n%d\n%d\n",ge,shi,bai);
9. if(i==ge*ge*ge+shi*shi*shi+bai*bai*bai)
10. printf("%d\n",i);
11. }

 

输出这个图像

 

1. int i,j,h;
2. for(i=1;i<7;i++)
3. {
4. for(j=i;j>=1;j--)
5. {
6. printf("_");
7. }
8. for(h=0;h<i;h++)
9. {
10. printf("%c",'F'-h);
11. }
12. putchar(10);
13. }
14. return0;
15. }

判断100以内质数 。1既不是质数也不是。。。。

1. int i ,j;
2. for(i=2;i<100;i++)
3. {
4. for(j=2;j<=i;j++)
5. {
6. if(( i%j==0)&&(i!=j))
8. break;
9. if(i==j)
11. printf("%d\n",i);
12. }
15. }

 计算某年的某天是第多少天。

1. # include <stdio.h>
2. int main(int argc,constchar*argv[])
3. {
4. int nian,yue,ri;
5. int flag =1;
6. int sum=0;
7. while(1)
8. {
9. puts("please input:");
11. if(scanf("%d %d %d",&nian,&yue,&ri)!=3)
12. {
13. puts("input error,try again:");
14. getchar();
15. getchar();
16. getchar();
17. continue;
19. }
20. if((nian<0||nian>=9999)||0>yue||yue>12)
21. {
23. puts("输入错误！请重新输入");
24. continue;
25. }
27. //判断输入到数字是否合理
28. if(yue==1||yue==3||yue==5||yue==7||yue==8||yue==10||yue==12)
29. {
30. if(0>ri||ri>31)
31. {
32. puts("日输入错误！请重新输入");
33. continue;
34. }
36. }
37. if(yue==4||yue==6||yue==9||yue==11)
38. {
39. if(ri>30)
40. {
41. puts("日输入错误！请重新输入");
42. continue;
43. }
44. }
46. {
47. if((nian%4==0)&&(nian%100!=0)||nian%400==0)
48. {
49. flag=1;
50. if(yue==2)
51. if(ri<0||ri>29)
52. {
53. puts("日输入错误！请重新输入");
54. continue;
55. }
57. }
58. else
59. {
60. if(yue==2)
61. if(ri<0||ri>28)
62. {
64. puts("日输入错误！请重新输入");
65. continue;
66. }
67. flag=0;
68. }
69. }
72. switch(yue-1)
73. {
75. case11:sum+=30;
76. case10:sum+=31;
77. case9:sum+=30;
78. case8:sum+=31;
79. case7:sum+=31;
80. case6:sum+=30;
81. case5:sum+=31;
82. case4:sum+=30;
83. case3:sum+=31;
84. case2:{if(flag)sum+=29;else sum+=28;}
85. case1:sum+=31;
87. }
89. printf("%d\n",sum+ri);
90. sum=0;
91. }
92. return0;
93. }