作业2017.10.12
来源:互联网 发布:现货倚天技术指标源码 编辑:程序博客网 时间:2024/05/22 13:34
#include<stdio.h>
int main()
{
int num=444,ge,shi,bai,qian,wan,sum;
printf("输入一个0-99999的整数");
scanf("%d",&num);
if(num>9999)
{sum=5;}
else if(num>999)
{sum=4;}
else if(num>99)
{sum=3;}
else if(num>9)
{sum=2;}
else
{sum=1;}
printf("输出位数:%d\n",sum);
printf("输出的每位数为");
wan=num/10000;
qian=(num-wan*10000)/1000;
bai=(num-wan*10000-qian*1000)/100;
shi=(num-wan*10000-qian*1000-bai*100)/10;
ge=(num-wan*10000-qian*1000-bai*100-shi*10);
switch(sum)
{
case 5:printf("%d,%d,%d,%d,%d\n",wan,qian,bai,shi,ge);
printf("反序为");
printf("%d,%d,%d,%d,%d\n",ge,shi,bai,qian,wan);
break;
case 4:printf("%d,%d,%d,%d\n",qian,bai,shi,ge);
printf("反序为");
printf("%d,%d,%d,%d\n",ge,shi,bai,qian);
break;
case 3:printf("%d,%d,%d\n",bai,shi,ge);
printf("反序为");
printf("%d,%d,%d\n",ge,shi,bai);
break;
case 2:printf("%d,%d\n",shi,ge);
printf("反序为");
printf("%d,%d\n",ge,shi);
break;
case 1:printf("%d\n",ge);
printf("反序为");
printf("%d\n",ge);
break;
}
return 0;
}
#include<stdio.h>
int main()
{
int li,jiangjin;
printf("输入利润:");
scanf("%d",&li);
if(li<=10000)
{jiangjin=li*0.1;}
else if(li>10000 && li<=20000)
{jiangjin=10000*0.1+(li-10000)*0.075;}
else if(li>20000 && li<=40000)
{jiangjin=10000*(0.1+0.075)+(li-20000)*0.05;}
else if(li>40000 && li<=60000)
{jiangjin=10000*(0.1+0.075)+20000*0.05+(li-40000)*0.03;}
else if(li>60000 && li<=100000)
{jiangjin=10000*(0.1+0.075)+20000*0.05+20000*0.03+(li-60000)*0.015;}
else
{jiangjin=10000*(0.1+0.075)+20000*0.05+20000*0.03+40000*0.015+(li-100000)*0.01;}
printf("奖金为:%d\n",jiangjin);
return 0;
}*/
#include <stdio.h>
int main()
{
int li;
double n1,n2,n4,n6,n10, j;
int a;
n1=10000*0.1;
n2=n1+10000*0.075;
n4=n2+20000*0.05;
n6=n4+20000*0.03;
n10=n6+40000*0.015;
printf("请输入利润li :");
scanf ("%d",&li);
a=li/10000;
if(a>10)
{a=10;}
switch(a)
{ case 0:j=li*0.1;break;
case 1:j=n1+(li-10000)*0.075;break;
case 2:
case 3:j=n2+(li-20000)*0.05;break;
case 4:
case 5:j=n4+(li-40000)*0.03;break;
case 6:
case 7:
case 8:
case 9:j=n6+(li-60000)*0.015;break;
case 10:j=n10+(li-100000)*0.01;break;
}
printf("奖金为:%6.2f\n",j);
return 0;
}
/*#include<stdio.h>
#include<math.h>
int main()
{ int h;
float x1=2,y1=2,x2=2,y2=-2,x3=-2,y3=2,x4=-2,y4=-2,x,y,a,b,c,d;
printf("请输入坐标点:") ;
scanf("%f,%f",&x,&y);
a=sqrt((x-x1)*(x*x1)+(y-y1)*(y-y1));
b=sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));
c=sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));
d=sqrt((x-x4)*(x-x4)+(y-y4)*(y-y4));
if(a>1 && b>1 && c>1 && d>1 )
{h=0;}
else
{h=10;}
printf("高度:%d\n",h);
return 0;
}*/
int main()
{
int num=444,ge,shi,bai,qian,wan,sum;
printf("输入一个0-99999的整数");
scanf("%d",&num);
if(num>9999)
{sum=5;}
else if(num>999)
{sum=4;}
else if(num>99)
{sum=3;}
else if(num>9)
{sum=2;}
else
{sum=1;}
printf("输出位数:%d\n",sum);
printf("输出的每位数为");
wan=num/10000;
qian=(num-wan*10000)/1000;
bai=(num-wan*10000-qian*1000)/100;
shi=(num-wan*10000-qian*1000-bai*100)/10;
ge=(num-wan*10000-qian*1000-bai*100-shi*10);
switch(sum)
{
case 5:printf("%d,%d,%d,%d,%d\n",wan,qian,bai,shi,ge);
printf("反序为");
printf("%d,%d,%d,%d,%d\n",ge,shi,bai,qian,wan);
break;
case 4:printf("%d,%d,%d,%d\n",qian,bai,shi,ge);
printf("反序为");
printf("%d,%d,%d,%d\n",ge,shi,bai,qian);
break;
case 3:printf("%d,%d,%d\n",bai,shi,ge);
printf("反序为");
printf("%d,%d,%d\n",ge,shi,bai);
break;
case 2:printf("%d,%d\n",shi,ge);
printf("反序为");
printf("%d,%d\n",ge,shi);
break;
case 1:printf("%d\n",ge);
printf("反序为");
printf("%d\n",ge);
break;
}
return 0;
}
#include<stdio.h>
int main()
{
int li,jiangjin;
printf("输入利润:");
scanf("%d",&li);
if(li<=10000)
{jiangjin=li*0.1;}
else if(li>10000 && li<=20000)
{jiangjin=10000*0.1+(li-10000)*0.075;}
else if(li>20000 && li<=40000)
{jiangjin=10000*(0.1+0.075)+(li-20000)*0.05;}
else if(li>40000 && li<=60000)
{jiangjin=10000*(0.1+0.075)+20000*0.05+(li-40000)*0.03;}
else if(li>60000 && li<=100000)
{jiangjin=10000*(0.1+0.075)+20000*0.05+20000*0.03+(li-60000)*0.015;}
else
{jiangjin=10000*(0.1+0.075)+20000*0.05+20000*0.03+40000*0.015+(li-100000)*0.01;}
printf("奖金为:%d\n",jiangjin);
return 0;
}*/
#include <stdio.h>
int main()
{
int li;
double n1,n2,n4,n6,n10, j;
int a;
n1=10000*0.1;
n2=n1+10000*0.075;
n4=n2+20000*0.05;
n6=n4+20000*0.03;
n10=n6+40000*0.015;
printf("请输入利润li :");
scanf ("%d",&li);
a=li/10000;
if(a>10)
{a=10;}
switch(a)
{ case 0:j=li*0.1;break;
case 1:j=n1+(li-10000)*0.075;break;
case 2:
case 3:j=n2+(li-20000)*0.05;break;
case 4:
case 5:j=n4+(li-40000)*0.03;break;
case 6:
case 7:
case 8:
case 9:j=n6+(li-60000)*0.015;break;
case 10:j=n10+(li-100000)*0.01;break;
}
printf("奖金为:%6.2f\n",j);
return 0;
}
/*#include<stdio.h>
#include<math.h>
int main()
{ int h;
float x1=2,y1=2,x2=2,y2=-2,x3=-2,y3=2,x4=-2,y4=-2,x,y,a,b,c,d;
printf("请输入坐标点:") ;
scanf("%f,%f",&x,&y);
a=sqrt((x-x1)*(x*x1)+(y-y1)*(y-y1));
b=sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));
c=sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));
d=sqrt((x-x4)*(x-x4)+(y-y4)*(y-y4));
if(a>1 && b>1 && c>1 && d>1 )
{h=0;}
else
{h=10;}
printf("高度:%d\n",h);
return 0;
}*/
阅读全文
0 0
- 作业2017.10.12
- [2017.10.6]作业01
- 2017.10.15作业2
- [2017.10.18]作业02
- 个人作业 2017.10.20
- [2017.10.23]作业04
- 2017.10.22作业
- 2017.10.31号作业
- 12-12作业
- 12-12作业
- 12-12作业
- 12周作业3
- 第12课堂作业
- 课堂作业12
- 9/12日作业
- unit 11-12作业
- 12天作业1
- python作业12
- JAVA工具类,Local解析
- 文章标题
- jQuery实现的截取字符串后面以省略号替代
- maven profile实现多环境打包
- mybatis延迟加载开关
- 作业2017.10.12
- 模板:块状链表
- springMVC中多个自定义拦截器方法的执行顺序
- elasticsearch+kibana安装
- Hive学习--架构和基本组成
- Linux下安装mysql
- 熵编码之CABAC
- 获取动态数据到echarts-地图
- Oracle tnsnames.ora和listener.ora