来自noj的期末复习
#include
using namespace std;
int main(){
int N,n,num=0;
cin>>N;
n=N;
while(N!=0){
num=num*10;
num+=N%10;
N/=10;
}
n==num?cout<<"yes"<<'\n':cout<<"no"<<endl;
return 0;
}
#include
#include
#include
using namespace std;
int main(){
double i,a=0;
cin>>i;
if (i<=10){
a=i*(0.1);
}else if(i<=20){
a=10*0.1+(i-10)*0.075;
}else if(i<=40){
a=10*0.1+10*0.075+(i-20)*0.05;
}else if(i<=60){
a=10*0.1+10*0.075+20*0.05+(i-40)*0.03;
}else if(i<=100){
a=10*0.1+10*0.075+20*0.05+20*0.03+(i-60)*0.015;
}else {a=10*0.1+10*0.075+20*0.05+20*0.03
+40*0.015+(i-100)*0.01;}
cout<<fixed<<setprecision(6)<<a<<endl;
return 0;
}
3.出租车费
#include
#include
#include
#include
using namespace std;
int main(){
double a,b,m;
cin>>a;
if(a<=2){
b=7;
}else if (a<=15){
m=ceil(a);//主要是学习ceil函数的应用
b=7+(m-2)*1.5;
}else {
m=ceil(a);
b=7+13*1.5+(m-15)*2.1;
}
cout<<fixed<<setprecision(6)<<b<<endl;
return 0;
}
#include
#include
#include
using namespace std;
int get_days(int year,int month){
switch(month){
case 1:case 3:case 5:case 7:
case 8:case 10:case 12:
return 31;
case 4:case 6:case 9:case 11:
return 30;
case 2:
if((year%4==0&&year%100!= 0)||year%400==0)
return 29;//闰年的概念
else return 28;
}
return 0;
}
int main(){
int year ,month,day;
scanf("%d-%d-%d",&year,&month,&day);
//记住scanf的用法
if (month <0||month>12)
return 1;
int days=day;
for (int i=1;i<month;i++){
int dm=get_days(year ,i);
days+=dm;
}
cout<<days<<endl;
return 0;
}
#include
#include
#include
using namespace std;
int main(){
float x,y;
int h;
scanf("%f,%f",&x,&y);
if((x-2)*(x-2)+(y-2)*(y-2)<=1||
(x-2)*(x-2)+(y+2)*(y+2)<=1||
(x+2)*(x+2)+(y-2)*(y-2)<=1||
(x+2)*(x+2)+(y+2)*(y+2)<=1){
h=10;
}else {h=0;}
cout<<h<<endl;
return 0;
}
#include
#include
#include
#include
using namespace std;
int main(){
int a=799,b,c,cnt=0,sum=0,d=1;
while(a>500&&a<800){
c=0;
b=a-1;
while(b>1){
if(a%b==0)
c=c+1;
b=b-1;
};
if(c==0){
cnt+=1;
sum+=d*a;
d=-1*d;
};
a=a-1;
}
cout<<cnt<<" "<<sum<<endl;
return 0;
}
7.你会存钱吗?
#include
#include
#include
#include
using namespace std;
int main(){
int i8,i5,i3,i2,i1,n8,n5,n3,n2,n1;
float max=0,term;
for(i8=0;i8<3;i8++)
for(i5=0;i5<=(20-8*i8)/5;i5++)
for(i3=0;i3<=(20-i8*8-i5*5)/3;i3++)
for(i2=0;i2<=(20-i8*8-i5*5-i3*3)/2;i2++){
i1=20-i8*8-i5*5-i3*3-i2*2;
term=2000.0*pow((double)(1+0.0063*12),(double)i1)
*pow((double)(1+0.0066*12),(double)i2)
*pow((double)(1+3*0.0069*12),(double)i3)
*pow((double)(1+5*0.0075*12),(double)i5)
*pow((double)(1+8*0.0084*12),(double)i8);
if(term>max){
max=term;
n1=i1;
n2=i2;
n3=i3;
n5=i5;
n8=i8;
}
}
cout<<n8<<" "<<n5<<" "<<n3<<" "<<n2<<" "<<n1<<endl;
cout<<max<<endl;
return 0;
}
//确实有点麻烦
还有好多题,我发现我好多不会。。。。