Accelerated C 習題 第3章

# include

# include
#include
#include
#include
# include
using
namespace std;
intmain()
typedef vector<
double
>
::size_type vec_sz;
vec_sz size = homework.
size()
;//判斷homework是否為空
if(size ==0)
//排序
sort
(homework.
begin()
, homework.
end())
;double median;
median = size %2==
0?(homework[size /2]
+ homework[size /2-
1])/
2: homework[size /2]
;//計算成績,設定精度
中值是乙個可將數值集合劃分為相等的上下兩部分的數值。對於有限的數集，通過把所有的值高低排序後找出正中間的乙個作為中位數。中間點取決於所有的數值。

由此可知，如果任何乙個我們已經讀到的值被丟棄，則會改變原先中間點的位置，這樣就得不出準確的中值。

#include

#include
#include
#include
using
namespace std;
intmain()
sort
(integers.
begin()
, integers.
end())
;typedef vector<
int>
::size_type vec_size;
vec_size size;
//集合總長度
size = integers.
size()
;//判斷是否為空
if(size ==0)
//集合中間位置
int mid = size /2;
//集合四分之一位置
}//不同的單詞if(
!same)
}for
(vec_size i =
0; i != words.
size()
;++i)
}

#include

< iostream>
#include
#include
#include
using
namespace std;
intmain()
//shortest_sz == 0這個一定要寫 不然 找不到最小的那個
if(shortest_sz ==
0|| str.
size()
< shortest_sz)
}	cout <<
"最長的字元："
<<  longest << endl;
cout << longest_sz << endl;
cout<<
"最短的字元："
				
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