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查看: 1199|回复: 9
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可以帮我看下..这个code,吗??
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#include<iostream>
using namespace std;
double Min(double,double,double);
double Max(double,double,double);
int main()
{
double no1, no2, no3, min, max;
cout<<"Enter 3 number \n";
cin>>no1>>no2>>no3;
min=Min(no1, no2 , no3);
cout<<"The smaller number is "<<min<<endl;
max=Max(no1, no2, no3);
cout<<"The largest number is "<<max<<endl;
return 0;
}
double Min(double no1, double no2, double no3)
{
if(no1<no2&&no1<no3)
return no1;
if(no2<no1&&no3<no2)
return no2;
else if(no3<no1&&no3<no2)
return no3;
else
cout<<"error"<<endl;
return 0;
}
double Max(double no1, double no2, double no3)
{
if(no1>no2&&no1>no3)
return no1;
if(no2>no1&&no3>no2)
return no2;
else if(no3>no1&&no3>no2)
return no3;
else
cout<<"error"<<endl;
return 0;
}
可以帮我看下..这个code,吗??
我的lagernumber 一直拿不到的比如1 2 3
他的lagernumber给我2 而不是 3
还有如果我要做成20组号码..有没有比较短的方法呢? |
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发表于 4-8-2007 10:31 PM
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double no1, no2, no3, min, max;
cout<<"Enter 3 number \n";
cin>>no1>>no2>>no3;
------------------------
这部分可以用array
double num[3];
cout <<"Enter 3 numbers\n";
for(int i=0;i<3;i++)
cin >>num;
-----------------------
然后那个
double min(double a,double b)
{
if(a>b)
return b;
else
return a;
}
--->call function时候, 要比较3个号码
--> min( num[0], min(num[1],num[2]) ); 可以这样 最里面的会return最小的号码
----------
max也一样咯 |
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发表于 5-8-2007 12:17 AM
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原帖由 tayar18 于 4-8-2007 09:59 PM 发表 
#include
using namespace std;
double Min(double,double,double);
double Max(double,double,double);
int main()
{
double no1, no2, no3, min, max;
coutno1>>no2>>no3;
min=Min(no1, no2 , ... -
- double Min(double no1, double no2, double no3)
- {
- double tmp = no1;
- if(no2 < tmp) tmp = no2;
- if(no3 < tmp) tmp = no3;
- return tmp;
- }
- double Max(double no1, double no2, double no3)
- {
- double tmp = no1;
- if(no2 > tmp) tmp = no2;
- if(no3 > tmp) tmp = no3;
- return tmp;
- }
复制代码 |
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楼主 |
发表于 5-8-2007 12:29 AM
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回复 #2 呵呵哟 的帖子
double min(double a,double b)
{
if(a>b)
return b;
else
return a;
}
这个......20个series也是酱吗?? |
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发表于 5-8-2007 01:16 AM
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原帖由 tayar18 于 5-8-2007 12:29 AM 发表 
double min(double a,double b)
{
if(a>b)
return b;
else
return a;
}
这个......20个series也是酱吗??
是不是20个号码,找出最小?
是的话
double num[20];
double mintmp;
for(int i=0;i<20;i++)
{
cout <<"enter num: ";
cin >> num;
}
mintmp = num[0]; //assume 1st num to be min
for(int i=0;i<20;i++)
{
mintmp = min(num,mintmp); //compare current num wif the stored min, if bigger , then store into mintmp, else, remain.
}
cout << "min = " <<mintmp; |
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发表于 5-8-2007 11:18 AM
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原帖由 呵呵哟 于 5-8-2007 01:16 AM 发表 
double num[20];
double mintmp;
for(int i=0;i<20;i++)
{
cout <<"enter num: ";
cin >> num;
}
mintmp = num[0]; //assume 1st num to be min
for(int i=0;i<20;i++)
{
mintmp = min(num[i ],mintmp); //compare current num wif the stored min, if bigger , then store into mintmp, else, remain.
}
cout << "min = " <<mintmp;
看tayar18写的program,我猜他应该还没有学到array吧,你写这些会不会把他弄乱啊??
[ 本帖最后由 hcong85 于 5-8-2007 11:32 AM 编辑 ] |
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楼主 |
发表于 5-8-2007 01:05 PM
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发表于 6-8-2007 03:38 AM
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发表于 14-8-2007 05:54 PM
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array比较effective&efficiency |
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发表于 23-8-2007 03:32 PM
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这个码运用了C++的 dynamic array. user 可以输入任何数目的号码,然后这个码可以显示最大和最小的号码。
#include <iostream>
using namespace std;
double Min(double [], int);
double Max(double [], int);
int main()
{
double *numbers;
int count;
double minNum, maxNum;
cout << "How many numbers you want to enter? ";
cin >> count;
numbers = new double[count];
cout << "Enter " << count << " numbers: ";
for(int x=0; x<count; x++)
{
cin >> numbers[x];
}
minNum = Min(numbers, count);
cout <<"The mininum number is " << minNum << endl;
maxNum = Max(numbers, count);
cout <<"The maximum number is " << maxNum << endl;
return 0;
}
double Min(double array[], int count)
{
double minNo;
minNo = array[0];
for(int x=1; x<count; x++)
{
if(minNo > array[x])
minNo = array[x];
}
return minNo;
}
double Max(double array[], int count)
{
double maxNo;
maxNo = array[0];
for(int x=1; x<count; x++)
{
if(maxNo < array[x])
maxNo = array[x];
}
return maxNo;
} |
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