jva 104 河洛之數

1) 矩陣內數字皆不可重複出現' e5 b3 X: }9 e% u& e0 N
(2) 程式每次執行，矩陣數字排列皆不相同
(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
{
public static void main(String arg[])% ?3 B0 _  i$ j* a+ l# _/ q
{$ k5 o! E; [/ Q+ b
      int num []={1,2,3,4,5,6,7,8,9};# f7 V: O$ L- g7 b6 X
      while(trun)
{& i3 _( t3 r5 |0 y: q$ b9 f
int r=(int)(Math.random()*9)+1;  R) J. f0 y, V3 H; h
int t=num[i];
num[i]=num[r];
num[r]=t;* p/ N- O% m; A2 F! {6 t3 H; v
int a1 = num[0] + num[1] + num[2] ;* ^; n  F7 p4 A/ b8 g
      int a2 = num[3] + num[4] + num[5] ;6 ]# J0 M( e  U; I/ [
      int a3 = num[6] + num[7] + num[8] ;6 U$ @" g/ W, u2 b& O+ o& d, ~
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;, t" j, w/ m! ]1 j; B/ y
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;) s% J. U& P4 Q- e; J
      int a8 = num[2] + num[4] + num[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;       ; g6 O  q5 W# M$ P
}
}

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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, K% O6 c5 m4 y' Y2 A
{: x8 T/ d, f4 H3 ^% w+ q& Y
public static void main(String arg[]). k8 G9 u0 E, w# h% W4 X: c' S- W
      {
         int num []={1,2,3,4,5,6,7,8,9};- e9 g9 x# }9 g: w
      while(trun)4 I) `6 A- W5 ]) D9 E1 b% {- z$ a
            {
            int r=(int)(Math.random()*9)+1;, l  {. \, M. u' H4 d( R
            int t=num[i];
            num[i]=num[r];# T* i. M1 o& C% |$ Z( |! O
            num[r]=t;5 {. w8 D, u6 D& B$ k$ s
            int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;; T# [9 v9 w+ H8 R- `
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;5 z, S0 }; M# m; F
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;) X6 m  ~* l0 e2 j5 s
      int a8 = num[2] + num[4] + num[6] ;2 ?: z: m3 p; b" E  h& u9 I2 A2 D
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          5 x6 l/ E' Y9 N3 u
}
}

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