jva 104 河洛之數

1) 矩陣內數字皆不可重複出現: b. y: ]2 N' v! ^
(2) 程式每次執行，矩陣數字排列皆不相同" X4 `1 B c0 g/ Z* B! G1 F. `' n4 i
(3) 矩陣之縱向、橫向、對角線數字總和皆為15

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

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2^#

johnson發表於 2011-12-31 11:35 | 只看該作者

public class j102
{. a! v7 M) T( j: A
public static void main(String arg[])0 U/ d& j* j8 ?$ d' ]1 _
{
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{- B: m8 s& [2 i
int r=(int)(Math.random()*9)+1;& S$ Q4 u+ a( s8 E$ C
int t=num[i];
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;. S1 [  Y" p$ b9 _  w
      int a2 = num[3] + num[4] + num[5] ;" A/ b3 X' O, ]2 Z6 G
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;" u( a- h) A  e6 @
      int a7 = num[0] + num[4] + num[8] ;* J4 o/ r& D1 b# m: |4 @5 l
      int a8 = num[2] + num[4] + num[6] ;6 W. u& x8 S$ x
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15); E- M9 x, Y6 d5 P. ?9 ?. s
      }break;
}
}

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3^#

kim發表於 2011-12-31 11:36 | 只看該作者

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

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4^#

eric5802發表於 2011-12-31 11:38 | 只看該作者

public class j102: n) i& l9 t3 u1 C! I$ G6 j8 }
{* J4 D. U* E$ _( _) ~4 m* Q9 ?* L  t
public static void main(String arg[])# Z/ U  `/ \& k$ I+ M
      {
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun); F5 }# I& ^# x2 O+ h1 }
            {
            int r=(int)(Math.random()*9)+1;
            int t=num[i];1 o9 k5 }8 a; O8 r
            num[i]=num[r];# c3 ]" y8 c; }5 F& J
            num[r]=t;  p) W9 \3 I7 T& q
            int a1 = num[0] + num[1] + num[2] ;! p3 P8 G( F6 {7 X/ y6 }
      int a2 = num[3] + num[4] + num[5] ;; {4 B% Q2 l! F. y& w6 t
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;3 H' _+ ~% ?) X& ?% g1 k1 j8 T
      int a5 = num[1] + num[4] + num[7] ;$ O/ m3 l- F; X
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;  D$ @  ]6 d  B% K9 \1 o
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)1 R3 R( w( z4 t1 u5 u! ]/ q
         }break;          6 l: o2 r  d# \7 X+ i3 _  ]  A
}
}

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