n x/2+n 0 x/2+n=x x=2*n=2*(0+ n)
n-1y/2+n-1 x y/2+n-1+x = y y = 2 *(x+n-1)
n-2z/2+n-2y z/2+n-2+y = z z = 2 *(y+n-2)
n-3 u/2+n-3 z u/2+n-3+z = u u = 2 *(z+n-3)
n-4v/2+n-4u v/2+n-4+u = v v = 2 *(u+n-4)
...
2 xx/2+2...
1 XXX/2+1 xx XXX/2+1+xx = XXX XXX = 2 *(xx+1)
您的代碼有問題,可以通過下面的代碼來驗證:
# include & ltstdio.h & gt
int main()
{
int n=0,x=0,I = 0;//x的初始值是0。
printf("輸入n:");
scanf("%d ",& ampn);
for(I = n;我& gt0;我-)
{
x =(x+I)* 2;
}
printf("猴子在第壹天得到了%d個桃子。\n”,x);
I = 0;
做{
int e;//吃了
i++;
e = x/2+I;
x-= e;//盈余
printf(" %-10de = %-10d r = %-10d \ n ",即x);
}
while(x & gt;0 ) ;
返回0;
}