C程式

#include <stdio.h>
int main()
{
int x=0,y=0,i;
for (i=1;i<5;i++)
{
if ((++x>=4)||(++y)>2)
{
x++;
}

printf("x=%d \n",x);
printf("y=%d \n",y);
printf("------------- \n");
}


return 0;
}

ANS：
x=1
y=1
-------------
x=2
y=2
-------------
x=4
y=3
-------------
x=6
y=3
-------------
i=1

x=1 (F) || y=1 (F)



i=2

x=2 (F) || y=2 (F)



i=3

x=3 (F) || y=3 (T)

-->x=4


i=4

x=5 (T)

成立，所以 || 後即不用執行

-->x=6

      y=3



++++++++++++++++++++++++++++++++++

#include <stdio.h>
int main()
{
int a=6,b=7;

printf("a=%d",a*=a--*--b);


return 0;
}

ANS：
a=180


a=a*(a--*--b)

  =5*(6*6)

  =180

++++++++++++++++++++++++++++++++++
#include <stdio.h>
int F(int n)
{
if(n==0) return 0;
if(n==1) return 1;

return F(n-1)+F(n-2);
}


int main()
{
printf("F(13)=%d",F(13));


return 0;
}
ANS：
F(13)=233


Fibonacci 數列

f(0)=0

f(1)=1

f(2)=1

f(3)=2

f(4)=3

f(5)=5

f(6)=8

f(7)=13

f(8)=21

f(9)=34

f(10)=55

f(11)=89

f(12)=144

f(13)=233

++++++++++++++++++++++++++++++++++
#include <iostream>
using namespace std;
int count=0;
int Get(int N,int M)
{
count++;
if (N==1) return (N != M ? 2:3);
else if (M==1) return 1;
else return Get(N-1,M)+Get(N-1,M-1);

}

int main()
{
int ans=Get(5,4);
cout<<"ans= "<<ans;
cout<<endl;
cout<<"count= "<<count;
}
ans= 32
count= 29