三道c语言的程序题要求要步骤全、最好把思路写上 先上200分在线等 C语言写程序运行结果题，求大神帮忙在线解答一下，写出解题过程...
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void main()//\u8ba1\u7b971 3 5\u7684\u9636\u4e58\u7684\u548c
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PS \u8fd9\u4e2a\u7a0b\u5e8f\u662f\u6709\u95ee\u9898\u7684\uff0c \u5224\u65ad\u4e2delse\u5e94\u8be5\u53bb\u6389\u3002 \u5426\u5219\u5728\u6781\u7aef\u60c5\u51b5\u4e0b\u4f1a\u51fa\u9519\u3002 \u4e0d\u8fc7\u5bf9\u4e8e\u8fd9\u4e2a\u6d4b\u8bd5\u7528\u4f8b\u4e0d\u4f1a\u3002
#include <stdio.h>
int main()
{
int a, b, opr;
printf("输入两个数字，如 3 5 （数字用空格分开）\n");
scanf("%d %d", &a, &b);
printf("选择计算方式：\n");
printf("1. 加， 2. 减， 3. 乘， 4. 除\n");
scanf("%d", &opr);
switch (opr)
{
case 1:
printf("%d + %d = %d\n", a, b, a+b);
break;
case 2:
printf("%d - %d = %d\n", a, b, a-b);
break;
case 3:
printf("%d X %d = %d\n", a, b, a*b);
break;
case 4:
if (b == 0)
{
printf("除数不能为 0\n");
return 1;
}
printf("%d / %d = %d\n", a, b, a/b);
break;
default:
printf("无效的计算方式\n");
return 1;
}
return 0;
}
第二题，输入购买册数m后，计算出总价 total，然后再根据total的大小，乘以对应的折扣率，即得最终价格。
#include <stdio.h>
int main()
{
unsigned int total, m, price;
price = 3; /* 3元每册 */
printf("要买多少册？\n");
scanf("%u", &m);
total = m*price; /* 总价 */
if (total > 20000)
total *= 0.75;
else if (total > 10000)
total *= 0.8;
else if (total > 2000)
total *= 0.85;
else if (total > 100)
total *= 0.9;
printf("应付 %u 元\n", total);
return 0;
}
第三题，让 x 从1开始循环到 9 ，依次代入 111*11* （10*x+1） 这个式子中，如果结果等于 111111 ，那么就找到对应的x了，然后打印结果即可。 最终结果 x = 9.
#include <stdio.h>
int main()
{
int x;
for (x = 1; x <= 9; x++)
{
if (111*11*(10*x + 1) == 111111)
{
printf("x = %d, 111111 = 111*11*%d1\n", x, x);
break;
}
}
return 0;
}
输入两个数字，然后做个菜选项，然后从键盘获取输入的数字，switch判断对应数字字符，然后进行运算，输出结果
#include <stdio.h>
int main()
{
int a,b;
char c;
printf("输入两个数字");
scanf("%d %d",&a,&b);
menu:
printf("%s\n","1.加");
printf("%s\n","2.减");
printf("%s\n","3.乘");
printf("%s\n","4.除");
while(1)
{
c=getchar();
switch(c)
{
case '1':
printf("a+b=%d\n",a+b);
goto menu;
break;
case '2':
printf("a-b=%d\n",a-b);
goto menu;
break;
case '3':
printf("a*b=%d\n",a*b);
goto menu;
break;
case '4':
if(b==0)
printf("除数不能为0\n");
else
printf("a/b=%d\n",a/b);
goto menu;
break;
}
}
}
第二题
输入顾客购书的数量，然后计算总额，比较之后根据要求，算出打着后的总金额，输出即可
#include <stdio.h>
#include <stdlib.h>
int main()
{
char num[100];
int a;
double money=0;
int pay=0;
while(1)
{
printf("输入购书册数：");
gets(num);
a=atol(num);
money=3*a;
if(money>=100 && money<2000)
money=money*0.90;
else if(money>=2000 && money<=10000)
money=money*0.85;
else if(money>10000 && money<20000)
money=money*0.80;
else if(money>=20000)
money=money*0.75;
printf("顾客应付%.2f元\n",money);
}
}
第三题，楼上的就行 上面程序中加上while循环，可循环输入显示，不必重新执行程序
#include <memory.h>
#include <stdlib.h>
#include <stdio.h>
#include "math.h"
bool CalReslut(double a,double b, int Operate,double& result){
bool bVal = true;
switch(Operate){
case 1:
result = a + b;
break;
case 2:
result = a - b;
break;
case 3:
result = a * b;
break;
case 4:
if(b == 0){
bVal = false;
printf("除数不能为0!");
break;
}
result = a / b;
break;
default:
printf("输入的操作运算符不对!");
bVal = false;
break;
}
return bVal;
}
float CalMenoy(int nNumber,float fPrice = 3.0f){
if(nNumber <= 0 || fPrice <= 0){
printf("请输入合法的数据!");
return 0;
}
float fVal = fPrice*nNumber;
if (fVal >= 20000.0f)
return fVal*0.75f;
else if(fVal >= 10000.0f)
return fVal*0.80f;
else if(fVal >= 2000.0f)
return fVal*0.85f;
else if(fVal >= 100.0f)
return fVal*0.90f;
else
return fVal;
}
bool GetX(char* pRightVal,int& x){
if(pRightVal == 0)
return false;
char val[2][5];
memset(val,0,sizeof(val));
char* p = pRightVal;
char* ps = p;
int i = 0;
int v = 0;
while (*(p++)){
if(v == 2)break;
i++;
if(*p == '*'){
memcpy(val[v],ps,i);
ps = ++p;
i =0 ;
v++;
}
}
if(v != 2)
return false;
int a = atoi(val[0]);
int b = atoi(val[1]);
x = (111111/(a*b)-1)/10;
return true;
}
void mainFace(){
printf("请输入以下字符：\n");
printf("1: 计算两数;\n");
printf("2: 计算图书总价\n");
printf("3: 计算111111等式中x的值\n");
printf("0: 退出\n");
}
void main()
{
while(1){
mainFace();
char a;
scanf("%c",&a);
switch(a){
case '0':
return;
case '1':
{
double x,y;
int operater;
double val = 0;
printf("请输入a,b的值以及操作符：\n");
printf("1: +;\n");
printf("2: -;\n");
printf("3: *;\n");
printf("4: \\;\n");
scanf("%lf%lf%d",&x,&y,&operater);
if(CalReslut(x,y, operater,val)){
printf("请算结果为：%lf\n\n",val);
}
}
break;
case '2':
{
int num;
printf("请输入图书的数量（价格为3元）：\n");
scanf("%d",&num);
float val = CalMenoy(num);
printf("应付：%f元\n\n",val);
}
break;
case '3':
{
char c[100];
memset(c,0,sizeof(c));
printf("请输入等式(如果111*11*)：\n");
scanf("%s",c);
int x;
if(GetX(c,x)){
printf("X = %d\n\n",x);
}
}
break;
default:
break;
}
}
}
1.void operation(int a,int b)
{
int oper;
printf("please input two integers");
scanf("%d%d",&a,&b);
printf("1.加2.减3.乘.4除\n");
scanf("%d",&oper);
switch(oper)
{
case(1):
printf("a+b=%d\n",a+b);
break;
case(2):
printf("a-b=%d\n",a-b);
break;
case(3):
printf("a*b=%d\n",a*b);
break;
case(4):
if(b==0) printf("除数不能为0");
else printf("a/b=%d\n",a/b);
break;
default:
printf("please input a number between 1and4\n");
}
}
2.float pay(int m)
{
float price=3*m;
if(price>20000) price*=0.75;
else if(price>10000) price*=0.8;
else if(price>2000) price*=0.85;
else if(price>100) price*=0.9;
return price;
}
1.if(button1.text="加")
{
textbox3.text=textbox1.text*textbox2.text
}
都是类似的 类型转换的话 我就不弄了
2.
现买3元的书m册
double a=double.pare(书的价格*m)
if(a>20000)
{
应付钱=a*0.75
}else
if(a>10000 && a<20000)
{
应付钱=a*0.8
}
这个是第二题的
可以不给分
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绛旓細// 绗竴棰榠nt main(){ int m[100] = {0}; int count = 0; printf("input the numbers:"); scanf("%d", &m[count]); while (m[count] != -1) { if (m[count] < 0 || m[count] > 4){ printf("input error!"); break; } count ++; ...
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绛旓細void main(){ int age;char sex;printf("please input your sex and age:");scanf("%c,%d",&sex,&age);if(sex=='M'){ if(age>=60) printf("you have retired!");else printf("you have not retired!");} else if(sex=='F'){ if(age>=55) printf("you have retired!");e...
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