1.&&和||的短路现象
2.条件 ？ 表达式1 ：表达式2 ；（条件为真时执行表达式1，否则执行表达式2）
作业：
1.铁路托运行李规定：行李重不超过50公斤的托运费按0.15元每公斤计算，超过50公斤的部分每公斤加收0.1元。
int main(int argc, const char * argv[])
{
float weight = 0,money = 0;
printf("输入货物重量：");
scanf("%f",&weight);
if (weight <= 50)
{
money = weight *
0.15;
}else if (weight >
50){
money = (weight - 50) * 0.25
+ 50 * 0.15;
}
printf("托运费为：%.2f\n",money);
return 0;
}
2.完成四则运算，如：用户输入34+56 输出90.00。用户输入两个数和运算符。通过运算符输出结果。
(1)
int main(int argc, const char * argv[])
{
float a = 0,b = 0;
char c =
‘\0‘;
printf("请输入计算内容：");
scanf("%f%c%f",&a,&c,&b);
if (c == ‘+‘)
{
printf("结果是：%.2f\n",a + b
);
}else if (c == ‘-‘)
{
printf("结果是：%.2f\n",a - b
);
}else if (c == ‘*‘)
{
printf("结果是：%.2f\n",a * b
);
}else if (c == ‘/‘)
{
printf("结果是：%.2f\n",a / b
);
}else{
printf("输入错误\n");
}
return 0;
}
(2)
int main(int argc, const char * argv[])
{
float a = 0,b = 0;
char c =
‘\0‘;
printf("输入计算内容：");
scanf("%f%c%f",&a,&c,&b);
switch (c)
{
case
‘+‘:
printf("结果是：%.2f\n",a +
b);
break;
case
‘-‘:
printf("结果是：%.2f\n",a -
b);
break;
case
‘*‘:
printf("结果是：%.2f\n",a *
b);
break;
case
‘/‘:
printf("结果是：%.2f\n",a /
b);
break;
default:
printf("输入错误\n");
break;
}
return 0;
}
3.输入三个数，求最大值和中间值。
(1)
int main(int argc, const char * argv[])
{
int a = 0,b = 0,c = 0,max = 0,mid =
0;
scanf("%d,%d,%d",&a,&b,&c);
max = a > b ? a
: b;
max = max > c ? max : c;
printf("最大值是：%d\n",max);
if ((a - c) * (a - b) <= 0)
{
mid = a;
}else if ((b - a) * (b - c) <=
0){
mid = b;
}else{
mid =
c;
}
printf("中间值是：%d\n",mid);
(2)
int a = 0,b = 0,c = 0,mid = 0;
printf("请输入三个数:");
scanf("%d,%d,%d",&a,&b,&c);
if (a >= b)
{
if (a >= c)
{
(b >=
c) ? (mid = b) : (mid = c);
}else
if (a < c){
mid =
a;
}
}else
if(a < b){
if (a >= c)
{
mid =
a;
}else if(a <
c){
(b
>= c) ? (mid = c) : (mid = b);
}
}
printf("中间数是：%d\n",mid);
(3)
int a = 0,b = 0,c = 0,mid =
0;
printf("请输入三个数:");
scanf("%d,%d,%d",&a,&b,&c);
if (a > b
&& a > c) {
mid = (b
> c) ? b : c;
}else if (b > c && b >
a){
mid = (a > c) ? a :
c;
}else{
mid = (a > b) ? a : b;
}
printf("中间值是：%d\n",mid);
(4)
int a = 0,b = 0,c = 0,mid = 0;
scanf("%d,%d,%d",&a,&b,&c);
if (a >= b
&& a <= c) {
mid =
a;
}else if (b >= a && b <=
c){
mid = b;
}else{
mid =
c;
}
printf("中间值是：%d\n",mid);
return 0;
}
4.有一个函数：x<1时，y=x；1<=x<10时，y=2x-1;x>=10时,y=3x-11;输入x,输出y。
int main(int argc, const char * argv[])
{
int x = 0,y = 0;
scanf("%d",&x);
if (x < 1)
{
y = x;
}else if (x >= 1 && x <=
10){
y = 2 * x -
1;
}else if(x >
10){
y = 3 * x -
11;
}
printf("y的值是：%d\n",y);
return 0;
}
5.给出一个百分制成绩，90分以上为‘A’，80~89为‘B’，70~79为‘C’，60~69为‘D’，60以下为‘E’。
int main(int argc, const char * argv[])
{
int x = 0;
scanf("%d",&x);
if (x >= 90 && x <= 100)
{
printf("A\n");
}else if (x >= 80 && x <=
89){
printf("B\n");
}else if (x >= 70 && x <=
79){
printf("C\n");
}else if (x >= 60 && x <=
69){
printf("D\n");
}else if (x >= 0 && x <=
59){
printf("E\n");
}else{
printf("输入错误\n");
}
return 0;
}
6.输入一个五位数，判断是否为回文数（如12321）
int main(int argc, const char * argv[])
{ int a = 0,wuwei = 0,siwei
= 0,sanwei = 0,erwei = 0,yiwei = 0;
printf("输入一个五位数：");
scanf("%d",&a);
wuwei = a /
10000;
siwei = (a - wuwei * 10000) /
1000;
sanwei = (a - wuwei * 10000 - siwei * 1000) /
100;
erwei = (a - wuwei * 10000 - siwei * 1000- sanwei *
100) / 10;
yiwei = a - wuwei * 10000 - siwei * 1000-
sanwei * 100 - erwei * 10;
if (wuwei == yiwei &&
siwei == erwei) {
printf("这个五位数是回文数\n");
}else{
printf("这个五位数不是回文数\n");
}
return
0;
}
原文：http://www.cnblogs.com/lxllanou/p/3576827.html