河內塔問題
#include<stdio.h>
#include<stdlib.h>
int fun_a(int)
void fun_b(int,int,int,int)
int main(void)
{
int n
int option
printf("題目二:河內塔問題\n")
printf("請輸入要搬移的圓盤數目\n")
scanf("％d",&n)
printf("最少搬移的次數為％d次\n",fun_a(n))
printf("是否顯示移動過程? 是請輸入1,否則輸入0\n")
scanf("％d",&option)
if(option==1)
{
fun_b(n,1,2,3)
}
system("pause")
return 0
}
int fun_a(int n)
{
int sum1=2,sum2=0,i
for(i=n i>1 i--)
{
sum1=sum1*2
}
sum2=sum1-1
return sum2
}
void fun_b(int n,int left,int mid,int right)
{
if(n==1)
printf("把第％d個盤子從第％d座塔移動到第％d座塔\n",n,left,right)
else
{
fun_b(n-1,left,right,mid)
printf("把第％d個盤子從第％d座塔移動到第％d座塔\n",n,left,right)
fun_b(n-1,mid,left,right)
}
}
標簽：
int
include
stdlib
fun_a
上傳時間：
2016-12-08
上傳用戶：努力努力再努力
function [U,center,result,w,obj_fcn]= fenlei(data)
[data_n,in_n] = size(data)
m= 2 ％ Exponent for U
max_iter = 100 ％ Max. iteration
min_impro =1e-5 ％ Min. improvement
c=3
[center, U, obj_fcn] = fcm(data, c)
for i=1:max_iter
if F(U)>0.98
break
else
w_new=eye(in_n,in_n)
center1=sum(center)/c
a=center1(1)./center1
deta=center-center1(ones(c,1),:)
w=sqrt(sum(deta.^2)).*a
for j=1:in_n
w_new(j,j)=w(j)
end
data1=data*w_new
[center, U, obj_fcn] = fcm(data1, c)
center=center./w(ones(c,1),:)
obj_fcn=obj_fcn/sum(w.^2)
end
end
display(i)
result=zeros(1,data_n) U_=max(U)
for i=1:data_n
for j=1:c
if U(j,i)==U_(i)
result(i)=j continue
end
end
end
標簽：
data
function
Exponent
obj_fcn
上傳時間：
2013-12-18
上傳用戶：ynzfm
