㈠ 求助,简单的C程序编写 商场购物,每满100元减20元,每满500,再减20,如购物770元,实付
#include<stdio.h>
int main(void)
{
int money;
printf("请输入购物消费金额.\n");
scanf("%d",&money);
int result=0;
result=money-money/100*20-money/500*20;
printf("实际付款金额为:%d",result);
printf("\n");
return 0;
}
这么简单的程序看得懂吧,不需要我说明吧
㈡ c语言买衣服最少用多少钱
#include<stdio.h>
#include<stdlib.h>
//衣服结构体
structclothes
{
floatfSellingPrice; //出售价格
floatfReturnPrice; //返还价格
};
int_tmain(intargc,_TCHAR*argv[])
{
intn=0;
printf("请输入购买衣服数量:");
scanf_s("%d",&n,sizeof(n));
//动态创建数量
clothes*p=newclothes[n];
inti=0;
intj=0;
for(i=0;i<n;i++)
{
printf("请输入第%d件衣服的价格和返还价格:",i+1);
scanf_s("%f,%f",&p[i].fSellingPrice,&p[i].fReturnPrice);
}
printf(" ");
floatfAllPrice=0; //总价格
floatfAllReturn=0; //总返现
//将其加和
for(i=0;i<n;i++)
{
//printf("%d,%.2f,%.2f ",i+1,p[i].fSellingPrice,p[i].fReturnPrice);
fAllPrice+=p[i].fSellingPrice;
fAllReturn+=p[i].fReturnPrice;
}
//总价格-总返现=最少需要花费价格
printf("最少需要花费价格为:%.2f ",fAllPrice-fAllReturn);
delete[]p;
p=NULL;
system("pause");
return0;
}
㈢ c语言换零钱
代码如下:
#include<stdio.h>
#include<stdlib.h>
intmain()
{
inti,j,k,n,cnt=0;
printf("Pleaseinputamountofmoney(unitjiao):");
scanf("%d",&n);
for(i=0;i<=(100/5);i++)
for(j=0;j<=(100/2);j++)
for(k=0;k<=100;k++)
if(i*5+j*2+k==n){
cnt++;
}
printf("Numberofconbinationsis:%d ",cnt);
system("pause");
return0;
}
运行结果:
㈣ C语言 零钱兑换
#include<stdio.h>
intmain()
{
intin;
scanf("%d",&in);
printf("%d",in/10);
in=in%10;
printf("%d",in/5);
in=in%5;
printf("%d ",in);
}
㈤ c语言 如何用switch语句编写一个有关商场购物金额优惠的选择程序
#include <stdio.h>
int main(int argc, char** argv)
{
int amount = 0;
scanf("%d", &amount);//输入顾客购买的总额
int status = amount/1000;
switch (status)
{
case 0: break;
case 1: amount = amount * 0.95; break;
case 2: amount = amount*0.90; break;
case 3: amount *= 0.85; break;
default: amount *= 0.80; break;
}
printf("%d\n", amount);//打印出打折后的总额
return 0;
}
㈥ c语言,从键盘输入一指定金额(以元为单位,如345.78),然后显示支付
a,b,c,d,e,f,g,h,i,j应为整型变量;角分处理做了点更改;
改后程序如下:
#include<stdio.h>
int main()
{
float num;
int a,b,c,d,e,f,g,h,i,j;
printf("请输入一个金额如345.78\n");
scanf("%f",&num);
a=(int)num/100;
b=(int)(num-a*100)/50;
c=(int)(num-a*100-b*50)/10;
d=(int)(num-a*100-b*50-c*10)/5;
e=(int)(num-a*100-b*50-c*10-d*5)/2;
f=(int)(num-a*100-b*50-c*10-d*5-e*2)/1;
g=(int)((num-a*100-b*50-c*10-d*5-e*2-f*1)*10)/5;
h=(int)((num-a*100-b*50-c*10-d*5-e*2-f*1-g*0.5)*10);
i=(int)((num-a*100-b*50-c*10-d*5-e*2-f*1-g*0.5-h*0.1)*20)/5;
j=(int)((num-a*100-b*50-c*10-d*5-e*2-f*1-g*0.5-h*0.1-i*0.05)*100);
printf("100元、50元、10元、5元、2元、1元、5角、1角、5分、1分各有%d,%d,%d,%d,%d,%d,%d,%d,%d,%d张",a,b,c,d,e,f,g,h,i,j);
return 0;
}