C语言第4天循环，流程控制。

时间：2023-03-09 15:27:35
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
C语言第4天循环，流程控制。
  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
C语言第4天循环，流程控制。
  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. }

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