工资

要求：求出不同范围的每月工资减去税后的实际所得。

include <stdio.h>

int main(){
int salary = 0; //记录工资
int pay = 0; //税
printf("请输入工资%d\n",salary);
scanf("%d",&salary);
//开始计算
if(salary < 5000){
pay = salary;
}else if(salary >= 5000 && salary < 8000){
pay = salary - (salary - 5000) * 0.1;
}else if(salary >= 8000 && salary < 10000){
pay = salary - (salary - 8000) * 0.15 - 3000 * 0.1;
}else if(salary >= 10000){
pay = salary - (salary - 10000) * 0.15 - 3000 * 0.1 - 2000*0.2;
}
printf("到手工资为：%d",pay);
return 0;
}

年龄转换实战

要求：把用阿拉伯数字表示的年龄转化为相对应的英文

include <stdio.h>

int main()
{
int age = 0;
//保存个位数
char *gewei[] = {"","one","two","three","four","five","six","seven","eight","nine"};
//保存10-19,""也是个字符串
char *temp[] = {"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
//保存十位
char *shiwei[] = {"twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};
printf("请输入年龄：");
scanf("%d",&age);
char *name;
if(age < 10){
name = gewei[age];
}
else if(age >= 10 && age <= 19){
int index = age % 10; //获取个位数
name = temp[index];
}
else (age >= 20); {
int g = age % 10;//获取个位数
int s = age / 10;//获取十位数
char * sString = shiwei[s-2];
char * gString = gewei[g];
printf("your age is %s-%s\n",sString,gString);//%s的s是String的s,
return 0;
}
printf("your age is %s\n",name);

return 0;

}

心得：
1.工资实战要求按要求将数据分类，并且运用if语句的嵌套循环来达成目的；
2.年龄转换实战给我的最大的收获就是能够使我将理论化为实践，且使我对数组的理解更加透彻。在修改并完善代码的过程中对如何写代码也有了新的体会，也就是写代码不要想得过多，先动手写，然后再边写边修改。
3.一定要注意写完一小段就要编译一下，出错就马上改。
4.要慢慢学会如何合理的简化程序
5.注意格式！方便检查。