Presentation is loading. Please wait.

Presentation is loading. Please wait.

第三次小考. #include using namespace std; int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<

Similar presentations


Presentation on theme: "第三次小考. #include using namespace std; int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<

1 第三次小考

2 #include using namespace std; int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } int main() { int ia[10]={0,2,3,6,8,6,12,28,16,11}; cout<

3 aaa(ib,0,9)

4 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=9 m=(0+9)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 4 a1=0 , a2=9  沒有 return 傳入 aaa(ib,0,4)

5 aaa(ib,0,9) aaa(ib,0,4)

6 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=4 m=(0+4)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 2 a1=0 , a2=4  沒有 return 傳入 aaa(ib,0,2)

7 aaa(ib,0, 9) aaa(ib,0, 4) aaa(ib,0, 2)

8 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=2 m=(0+2)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 1 a1=0 , a2=2  沒有 return 傳入 aaa(ib,0,1)

9 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1)

10 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=1 m=(0+1)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 0 a1=0 , a2=1  沒有 return 傳入 aaa(ib,0,0)

11 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) aaa(ib,0,0)

12 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=0 m=(0+2)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 1 a1=0 , a2=0  return ib[0]

13 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) aaa(ib,0,0) return ib[0] (ib[0]=0) u=aaa(ib,0,0)=0

14 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=1 m=(0+1)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 0 a1=0 , a2=1  沒有 return aaa(ib,0,0) return ib[0] 所以 u=ib[0] (u=0) u=0 傳入 aaa(ib,1,1)

15 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) aaa(ib,0,0) u=0 return ib[0] (ib[0]=0) aaa(ib,1,1)

16 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=1 , a2=1 m=(1+1)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 1 a1=a2=1  return ib[1]

17 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) aaa(ib,0,0) u=0 return ib[0] (ib[0]=0) aaa(ib,1,1) v=2 return ib[1] (ib[1]=2) v=aaa(ib,1,1)=2

18 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=1 m=(0+1)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 0 a1=0 , a2=1  沒有 return 傳入 aaa(ib,0,0) u = 0 v = aaa(ib,1,1) v = 2 u=0,v = 2 v>u return v (v=2)

19 v>u return v (v=2) aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) u=2 aaa(ib,0,0) u=0 return ib[0] (ib[0]=0) aaa(ib,1,1) v=2 return ib[1] (ib[1]=2)

20 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=2 m=(0+2)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 1 a1=0 , a2=2  沒有 return u=aaa(ib,0,1)=2 u=2 傳入 aaa(ib,2,2)

21 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) u=2 aaa(ib,0,0) u=0 return ib[0] (ib[0]=0) aaa(ib,1,1) v=2 return ib[1] (ib[1]=2) aaa(ib,2,2)

22 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=2 , a2=2 m=(2+2)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 2 a1=a2=2  return ib[2] (ib[2]=3)

23 aaa(ib,0,9) aaa(ib,0,4) aaa(ib,0,2) aaa(ib,0,1) u=2 aaa(ib,0,0) u=0 return ib[0] (ib[0]=0) aaa(ib,1,1) v=2 return ib[1] (ib[1]=2) aaa(ib,2,2) v=3 return ib[2] (ib[2]=3) v=aaa(ib,2,2)=3

24 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=2 m=(0+2)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 1 a1=0 , a2=2  沒有 return u=aaa(ib,0,1)=2 u=2 傳入 aaa(ib,2,2) v=3 u=2,v = 3 v>u return v (v=3)

25 v>u return v (v=3)

26 int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<v)return u;else return v; } 傳入參數 a1=0 , a2=4 m=(0+4)/2 …… 因為 m 屬於 int ,所以 m 等於整數部份 m = 2 a1=0 , a2=4  沒有 return 傳入 aaa(ib,0,2) aaa(ib,0,2) = 3 u = 3 傳入 aaa(ib,3,4)

27 v>u return v (v=3)

28 加油 ~~!! 繼續做下去一定會是 以下結果

29 aaa(ib,0,9) 28 aaa(ib,0,4) u=8 aaa(ib,0,2) u=3 aaa(ib,0,1) u=2 aaa(ib,0,0) u=0 return ib[0] aaa(ib,1,1) v=2 return ib[1] aaa(ib,2,2) v=3 return ib[2] aaa(ib,3,4) v=8 aaa(ib,3,3) u=6 return ib[3] aaa(ib,4,4) v=8 return ib[4] aaa(ib.5,9) v=28

30 aaa(ib,0,9) 28 aaa(ib,0,4) u=8 aaa(ib.5,9) v=28 aaa(ib,5,7) u=28 aaa(ib,5,6) u=12 aaa(ib,5,5) u=6 return ib[5] aaa(ib,6,6) v=12 return ib[6] aaa(ib,6,7) v=28 aaa(ib,6,6) u=12 return ib[6] aaa(ib,7,7) v=28 return ib[7] aaa{ib,8,9) v=16 aaa(ib,8,8) u=16 return ib[8] Aaa(ib,9,9) v=11 return ib[9]


Download ppt "第三次小考. #include using namespace std; int aaa(int *ib,int a1,int a2) { int u,v; int m=(a1+a2)/2; if(a1==a2)return ib[a1]; u=aaa(ib,a1,m); cout<<"u="<

Similar presentations


Ads by Google