printf("max=%d\n",c);
else
printf("max=%d\n",a);
}
方法二:
#include
void main()
{
int a,b,c,temp,max;
printf("请输入三个整数:
");
scanf("%d,%d,%d",&a,&b,&c);
temp=(a>b)?
a:
b;
max=(temp>c)?
temp:
c;
printf("最大数是%d\n",max);
}
5.5#includevoidmain(){intx,y;scanf("%d",&x);if(x<1){y=x;printf("x=%d,y=x=%d\n");}elseif(x<10){y=2*x-1;printf("x=%d,y=2*x-1=%d\n",x,y);}else{y=3*x-11;printf("x=%d,y=3*x-11=%d\n",x,y);}}
5.6#includevoidmain(){floatscore;chargrade;printf("请输入学生成绩");scanf("%f",&score);while(score>100||score<0){printf("\n输入有误,请重输");scanf("%f",&score);}switch((int)(score/10)){case10:
case9:
grade='A';break;case8:
grade='B';break;case7:
grade='C';break;case6:
grade='D';break;case5:
case4:
case3:
case2:
case1:
case0:
grade='E';}printf("成绩%.1f,相应等级%c\n",score,grade);}5.7#include
void main()
{
long int num;
int indiv,ten,hundred,thousand,ten_thousand,place;
printf("请输入一个整数:
");
scanf("%ld",&num);
if(num>9999)
place=5;
else if(num>999)
place=4;
else if(num>99)
place=3;
else if(num>9)
place=2;
else place=1;
printf("place=%d\n",place);
printf("每位数字为:
");
ten_thousand=num/10000;
thousand=(int)(num-ten_thousand*10000)/1000;
hundred=(int)(num-ten_thousand*10000-thousand*1000)/100;
ten=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10;
indiv=(int)(num-ten_thousand*10000-thousand*1000-hundred*100-ten*10);
switch(place)
{
case 5:
printf("%d,%d,%d,%d,%d",ten_thousand,thousand,hundred,ten,indiv);
printf("\n反序数字为:
");
printf("%d,%d,%d,%d,%d",indiv,ten,hundred,thousand,ten_thousand);break;
case 4:
printf("%d,%d,%d,%d",thousand,hundred,ten,indiv);
printf("\n反序数字为:
");
printf("%d,%d,%d,%d",indiv,ten,hundred,thousand);break;
case 3:
printf("%d,%d,%d",hundred,ten,indiv);
printf("\n反序数字为:
");
printf("%d,%d,%d",indiv,ten,hundred);break;
case 2:
printf("%d,%d",ten,indiv);
printf("\n反序数字为:
");
printf("%d,%d",indiv,ten);break;
case 1:
printf("%d",indiv);
printf("\n反序数字为:
");
printf("%d",indiv);break;
}
}
5.8#include
void main()
{
long i;
float bonus,bon1,bon2,bon4,bon6,bon10;
bon1=100000*0.1;
bon2=bon1+100000*0.075;
bon4=bon2+200000*0.05;
bon6=bon4+200000*0.03;
bon10=bon6+400000*0.015;
scanf("%ld",&i);
if(i<=100000)
bonus=i*0.1;
else if(i<=200000)
bonus=bon1+(i-100000)*0.075;
else if(i<=400000)
bonus=bon2+(i-200000)*0.05;
else if(i<=600000)
bonus=bon6+(i-600000)*0.015;
else
bonus=bon10+(i-1000000)*0.01;
printf("奖金是%10.2f",bonus);
}
法二:
5.8#include
void main()
{
long i;
float bonus,bon1,bon2,bon4,bon6,bon10;
int c;
bon1=
bon2=
bon4=
bon6=
bon10=
printf("请输入利润i:
");
scanf("%ld",&i);
c=i/100000;
if(c>10)
c=10;
switch(c)
{
case 0:
bonus=100000*0.1;break;
case 1:
bonus=bon1+(i-100000)*0.075;break;
case 2:
case 3:
bonus=bon2+(i-200000)*0.05;break;
case 4:
case 5:
bonus=bon4+(i-400000)*0.03;break;
case 6:
case 7:
case 8:
case 9:
bonus=bon6+(i-600000)*0.015;break;
case 10:
bonus=bon10+(i-1000000)*0.01;
}
printf("奖金是%.2f",bonus);
}
6.1#include
voidmain()
{
#include
voidmain()
{
intp,r,n,m,t;
scanf("%d,%d",&m,&n);
if(m {
t=m;m=n;n=t;
}
p=m*n;
while(n!
=0)
{
r=m%n;
m=n;
n=r;
}
printf("最大公约数为:
%d\n",m);
printf("最小公倍数为:
%d\n",p/m);
}
6.2 #include
voidmain()
{
inta,n,i;
longintsum=0,t;
scanf("%d%d",&a,&n);
i=1;
while(i<=n)
{
t=a*(long)(pow(10,i)-1)/9;
sum+=t;
i++;
}
printf("sum=%ld\n",sum);
}
6.3#includevoidmain(){inta,n,i=1,sn=0,t;printf("a,n=:
");scanf("%d,%d",&a,&n);t=a;while(i<=n){sn=sn+t;t=t*10+a;i++;}printf("a+aa+aaa+......=%d\n",sn);}6.4#include
void main()
{
float s=0,t=1;
int n;
for(n=1;n<=20;n++)
{
t=t*n;
s=s+t;
}
printf("1!
+2!
+...+20!
=%f\n",s);
}
6.5#includevoidmain(){intn1=100,n2=50,n3=10;floatk;floats1=0,s2=0,s3=0;for(k=1;k<=n1;k++)s1=s1+k;for(k=1;k<=n2;k++)s2=s2+k*k;for(k=1;k<=n3;k++)s3=s3+1/k;printf("总和=%8.2f\n",s1+s2+s3);}6.6#includevoidmain(){inti,j,k,n;printf("水仙花数是:
");for(n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if(n==i*i*i+j*j*j+k*k*k)printf("%4d",n);}}
6.9
#include
voidmain()
{
floatsn=100,hn=sn/2;intn;
for(n=2;n<=10;n++)
{
sn=sn+2*hn;
hn=hn/2;
}
printf("第10次落地时共经过%fm.\n",sn);
printf("第10次反弹%fm.\n",hn);
}
6.10
#include
voidmain()
{
intday,x1,x2;
day=9;x2=1;
while(day>0)
{
x1=(x2+1)*2;
x2=x1;
day--;
}
printf("total=%d\n",x1);
}
6.14
#include
voidmain()
{
inti,j,k;
for(i=0;i<=3;i++)
{
for(j=0;j<=2-i;j++)
printf("");
for(k=0;k<=2*i;k++)
printf("*");
printf("\n");
}
for(i=0;i<=2;i++)
{
for(j=0;j<=i;j++)
printf("");
for(k=0;k<=4-2*i;k++)
printf("*");
printf("\n");
}
}