解题基础:
scanf("输入模式",地址列表);//输入语句及其格式。注:地址列表一定要加取地址符号&,如:&a,&b
printf("输出模式",输出列表);//输出语句及其格式。注:输出模式和输出列表一一对应,如:printf("%d%d",a,b);
switch(表达式){
case 常量值1:
若干语句1
break;//可省略
case 常量值2:
若干语句2
break;//可省略
......
case 常量值n:
若干语句n
break;//可省略
default://可省略
若干语句
}//开关语句及其格式,若表达式的值等于某个常量值,则进行某常量值相对应的语句,若没遇到break,则接着运行下一个常量值后面的语句,直到遇到break为止,若表达式的值不等于某个常量值,则执行default后面的若干语句,default可省略(不执行语句)
思路:因为题目要求,需要一个输入的值和一个输出的值,因为定义a的是一个整型变量int,所以a/100000的值也是一个int型变量(整数)(注:C语言的取整不是四舍五入,是直接舍去小数),直接用switch函数可以写出运算语句(当然if也行,不过比较麻烦),t=对应区间的全部数*相应利润+(a-区间最大数)*超出相应的利润
答案:
#include<stdio.h> int main(){ int a,t; scanf("%d",&a); switch(a/100000){ case 0: t=a*0.1; break; case 1: t=100000*0.1+(a-100000)*0.075; break; case 2: case 3: t=100000*0.1+100000*0.075+(a-200000)*0.05; break; case 4: case 5: t=100000*0.1+100000*0.075+200000*0.05+(a-400000)*0.03; case 6: case 7: case 8: case 9: t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(a-600000)*0.015; default: t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(a-1000000)*0.01; } printf("%d",t); return 0; }
ps:当a/100000=2时,因为常量值后面的若干语句中没有break,则接着运行下一个常量值(case 3)后面的若干语句,直到遇到break为止
同理,后面的4,6,7,8都是一样的
程序结束return 0;不可省。
因为题目表达原因,生活中t应该为浮点型常量,保留两位小数,可是答案不给对,所以代码就以答案为准
若对switch函数还有疑问,可以对应我的上一个题解题解1008:C语言程序设计教程(第三版)课后习题5.6 (C语言描述)参照学习
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#include<stdio.h> int main() { int a,b; scanf("%d",&a); switch(a/100000){ case 0: b=a*0.1;break; case 1: b=100000*0.1+(a-100000)*0.075;break; case 2: case 3: b=100000*0.1+100000*0.075+(a-200000)*0.05;break; case 4: case 5: b=100000*0.1+100000*0.075+200000*0.05+(a-400000)*0.03;break; case 6: case 7: case 8: case 9: b=100000*0.1+100000*0.075+200000*0.05+400000*0.03+(a-600000)*0.015; default: b=100000*0.1+100000*0.075+200000*0.05+400000*0.03+600000*0.015+(a-1000000)*0.001; } peintf("%d",b); return 0; } 这个哪错了
#include<stdio.h> int main() { int amount, profit; scanf("%d",&amount); profit = 0; if(amount > 1000000){ profit = (amount -1000000)*0.01; amount -= amount - 1000000; } if(amount > 600000){ profit += (amount - 600000) * 0.015; amount -= amount - 600000; } if(amount > 400000){ profit += (amount - 400000) * 0.03; amount -= amount-400000; } if(amount > 200000){ profit += (amount - 200000 ) * 0.05; amount -= amount-200000; } if(amount > 100000){ profit += (amount - 100000 ) * 0.75; amount -= amount-100000; } profit += amount * 0.1; printf("%d\n", profit); re
#include <stdio.h> int main() { int profit=0; int bonus,b1=0,b2=0,b3=0,b4=0,b5=0,b6=0; scanf("%d",&profit); if(profit>1000000) { b6=(profit-1000000)*0.01; profit = 1000000; }; if(profit<=1000000 && profit>600000){ b5=(profit-600000)*0.015; profit = 600000; }; if(profit<=600000 && profit>400000){ b4=(profit-400000)*0.03; profit=400000; }; if(profit<=400000 && profit>200000){ b3=(profit-200000)*0.05; profit=200000; }; if(profit<=200000 && profit>100000){ b2=(profit-100000)*0.075; profit=100000; }; if(profit<=100000){ b
#include <stdio.h> int main() { int profit=0; int bonus,b1=0,b2=0,b3=0,b4=0,b5=0,b6=0; scanf("%d",&profit); if(profit>1000000) { b6=(profit-1000000)*0.01; profit = 1000000; }; if(profit<=1000000 && profit>600000){ b5=(profit-600000)*0.015; profit = 600000; }; if(profit<=600000 && profit>400000){ b4=(profit-400000)*0.03; profit=400000; }; if(profit<=400000 && profit>200000){ b3=(profit-200000)*0.05; profit=200000; }; if(profit<=200000 && profit>100000){ b2=(profit-100000)*0.075; profit=100000; }; if(profit<=100000){ b
#include<stdio.h> int main() { int i,t; scanf("%d",&i); if(i<=100000) { t=i*0.1; } else if(i>100000 && i<=200000) { t=100000*0.1+(i-100000)*0.075; } else if(i>200000 && i<=400000) { t=100000*0.1+100000*0.075+(i-200000)*0.05; } else if(i>400000 && i<=600000) { t=100000*0.1+100000*0.075+200000*0.05+(i-400000)*0.03; } else if(i>600000 && i<=1000000) { t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+(i-600000)*0.015; } else if (i>1000000) { t=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(i-1000000)*0.01; } printf("%d",t); return 0; } 这个哪里错了
#include <stdio.h> #include <stdlib.h> int a,b;//a存利润,b算钱 main() { scanf("%d",&a); if(a<=100000) b=0.1*a; else if(a<=200000&&a>100000)//&&a>100000 b=10000+(a-100000)*0.075; else if(a<=400000&&a>200000) b=10000+100000*0.075+(a-200000)*0.05; else if(a<=600000&&a>400000) b=10000+100000*0.075+200000*0.05+(a-400000)*0.03; else if(a<=1000000&&a>600000) b=10000+100000*0.075+200000*0.05+200000*0.03+(a-600000)*0.015; else if(a>1000000) b=10000+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(a-1000000)*0.01; printf("%d",b); }
#include<stdio.h> int main() { int x,y; scanf("%d",&x); if(x<=100000) y=x/10; else if(x>100000 && x<=200000) y=100000*0.1+(x-100000)*0.075; else if(x>200000 && x<=400000) y=100000*0.1+100000*0.075+(x-200000)*0.05; else if(x>400000 && x<=600000) y=100000*0.1+100000*0.075+200000*0.05+(x-400000)*0.03; else if(x>600000 && x<=1000000) y=100000*0.1+100000*0.075+200000*0.05+200000*0.03; else y=100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(x-1000000)*0.01; printf("%d\n,"y); return 0; } 这个哪里不对了呢
没学好c语言 2019-04-22 17:16:31 |
printf("%d",b);
给我五美元 2019-07-23 16:34:26 |
0.001? printf (疑问:还有为什么在 case 9 后没有break呢)