Question

C++的几道编程题，麻烦高手一一解答，100分。另求C++考试重点

1.编写程序，用户输入一些整数，该程序分别计算出所有奇数和所有偶数之和。
2.编写程序求标准一元二次方程的解 ax^2+bx+c=0 由用户输入abc的值
3.编写程序 用循环语句打印如下图案，用户决定图案行数
@
@#@
@###@
@#####@
@###@
@#@
@

4.编写程序，输入年月日信息，并输出这一天为这一年的第几天
5.编程求鸡兔同笼问题，若鸡兔共100只脚，利用循环计算鸡兔各几只

另求C++考试重点~~
可以发到我邮箱~~ftrain@vip.qq.com
在此谢过~~~分我有很多~~~可以给你加滴

Answer 1

1:
void main()
{
int a[10],i,m=0,n=0;
cout<<"请输入10个整数:"<<endl;
for(i=0;i<10;i++)
{
cin>>a[i];
if(a[i]%2==0)
m+=a[i];
else
n+=a[i];
}
cout<<"奇数和为:"<<m<<",偶数和为:"<<n<<endl;
}
2:
void main()
{
double a,b,c,d,x,x1,x2;
cout<<"Input values of a,b,and c: \n";
cin>>a>>b>>c;
d=b*b-4*a*c;
if(a==0 && b==0)
{
cout<<"\nroots are imaginary"<<endl;
}
else
{
if(a==0)
{
x=-c/b;
cout<<"x="<<x<<endl;
}
else
{
if(d<0)
{
cout<<"\nroots are imaginary"<<endl;
}
else
if(d==0)
{
x=(-b)/(2.0*a);
cout<<"x="<<x<<endl;
}
else
{
x1=(-b+sqrt(d))/(2.0*a);
x2=(-b-sqrt(d))/(2.0*a);
cout<<"x1="<<x1<<",x2="<<x2<<endl;
}
}
}
}
3:
#include <iostream>
using namespace std;
void main()
{
int h,i,j;
cin>>h;
for(i=0;i<h;i++)
{
for(j=0;j<h-i-1;j++)
cout<<" ";
cout<<"@";
for(j=0;j<2*i-1;j++)
cout<<"#";
if(i!=0)
cout<<"@";
cout<<endl;
}
for(i=h-2;i>=0;i--)
{
for(j=0;j<h-i-1;j++)
cout<<" ";
cout<<"@";
for(j=0;j<2*i-1;j++)
cout<<"#";
if(i!=0)
cout<<"@";
cout<<endl;
}
}
4:
#include<iostream>
using namespace std;
bool leapyear(int a) //判断是否为闰年
{
return ((a%4==0 && a%100!=0)||a%400==0);
}
int monthdays(int a,int b) //计算一个月的总天数
{
if(b==1||b==3||b==5||b==7||b==8||b==10||b==12)
return 31;
if(b==4||b==6||b==9||b==11)
return 30;
if(b==2)
if(leapyear(a))
return 29;
else
return 28;
return 0;
}
int yeardays(int a,int b,int c) //计算当前日期为当年的第几天
{
int i,days=0;
for(i=1;i<b;i++)
days+=monthdays(a,i);
return days+c;
}
void main()
{
int y,m,d,days;
cin>>y>>m>>d;
days=yeardays(y,m,d);
cout<<"这是第"<<days<<"天"<<endl;
}

5:
题目不全。头多少个？
只这样算答案很多：
void main()
{
int a,b;
for(a=1;a<=100/2;a++)
for(b=1;b<=100/4;b++)
if((2*a+4*b)==100)
cout<<"鸡"<<a<<"只,兔子"<<b<<"只"<<endl;
}

Answer 2

试试。
1.编写程序，用户输入一些整数，该程序分别计算出所有奇数和所有偶数之和。

#include <iostream.h>

main()
{
int n,sum=0,sum1=0;
cin>>n;//输入一个整数
for(int i=1;i<=n;i++)
{
if(i%2!=0) //奇数相加
sum=sum+i;
else
sum1=sum1+i;
}
cout<<n<<"的奇数相加="<<sum<<" "<<n<<"的偶数相加="<<sum1<<endl;
}

4.编写程序，输入年月日信息，并输出这一天为这一年的第几天
#include "iostream.h"
int main()
{
int year,month,date,numOfDay=0;
cout<<"请输入日期:";
cin>>year>>month>>date;
switch(month)
{
case 12:
numOfDay=numOfDay+30;
case 11:
numOfDay=numOfDay+31
case 10:
numOfDay=numOfDay+30;
case 9:
numOfDay=numOfDay+31;
case 8:
numOfDay=numOfDay+31;
case 7:
numOfDay=numOfDay+30
case 6:
numOfDay=numOfDay+31;
case 5:
numOfDay=numOfDay+30;
case 4:
numOfDay=numOfDay+31;
case 3:
numOfDay=numOfDay+((year%4==0&&year%100!=0)||(year%400==0)?29:28);
case 2:
numOfDay=numOfDay+31;
case 1:
numOfDay=numOfDay+date;
cout<<year<<"年"<<month<<"月"<<date<<"日是"<<year<<"年的第"<<numOfDay<<"天."<<endl;
break;
default:
cout<<"出错啦！"<<endl;
}
return 0;
}
VC6编写通过。

Answer 3

#include <iostream>
using namespace std;

void main()
{
int length;
int *data;
int sumforodd=0;
int sumforeven=0;
cout<<"请输入数列长度"<<endl;
cin>>length;
if(length>0?(length=1)-1:length)data = new int[length];
else data = new int ;
cout<<"请输入整数,回车结束输入，数字之间输入空格"<<endl;
for(int i=0;i<length;i++)
{
cin>>data[i];
(data[i]%2==1)?(sumforodd+=data[i]):(sumforeven+=data[i]);
}
cout<<"Sum of odds: "<<sumforodd<<endl;
cout<<"Sum of evens: "<<sumforeven<<endl;
}
这个是第一题的答案，有空再帮你写其他题目~~

Answer 4

3
#include <iostream>

using namespace std;

int main(int argc, char** argv)
{
int i,j;
char map[7][7] = {
{' ',' ',' ','@',' ',' ',' '},
{' ',' ','@','#','@',' ',' '},
{' ','@','#','#','#','@',' '},
{'@','#','#','#','#','#','@'},
{' ','@','#','#','#','@',' '},
{' ',' ','@','#','@',' ',' '},
{' ',' ',' ','@',' ',' ',' '}
};
for(i=0;i<7;++i){
for(j=0;j<7;++j)
cout<<map[i][j];
cout<<endl;
}
return 0;
}

Answer 5

5道题打不下，你能留一下能联系的方式吗？

追问

邮箱~~ftrain@vip.qq.com

追答

3.
#include
#include
using namespace std;

char c[10][10];

int main() {
int delta = 2;
int left, right;
int n = 1;
while (1) {
left = 3 - n / 2;
right = 3 + n / 2;
for (int i = 0; i left)
cout
using namespace std;
int main() {
int year, month, day;
cout > year >> month >> day;
int m = 1;
int d = 0;
switch(m) {
case 1 : if (month == 1) d += day; else d += 31;
case 2 : if (month == 2) d += day; else if ((year % 4 == 0 && year % 100 != 0)||year % 400 == 0) d += 29; else d += 28;
case 3 : if (month == 3) d += day; else d += 31;
case 4 : if (month == 4) d += day; else d += 30;
case 5 : if (month == 5) d += day; else d += 31;
case 6 : if (month == 6) d += day; else d += 30;
case 7 : if (month == 7) d += day; else d += 31;
case 8 : if (month == 8) d += day; else d += 31;
case 9 : if (month == 9) d += day; else d += 30;
case 10 : if (month == 10) d += day; else d += 31;
case 11 : if (month == 11) d += day; else d += 30;
case 12 : if (month == 12) d += day; else d += 31;
}
cout
using namespace std;
int main() {
int n = 100;
int i, j;
for (i = 0; i <= n / 4; i++) {
if ((100 - i * 4) % 2 == 0) {
cout << "rabbit num is " << i << "chicken num is " << (100 - i * 4) / 2 << endl;
}
}
system("pause");
}