jva 104 河洛之數

1) 矩陣內數字皆不可重複出現! U: R! f) E8 b2 P! d" @; d+ @
(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 j1021 \  J) E7 z/ f& n, S8 N) d% x- F
{' S; @4 N  ]* a8 F; y% z2 p
public static void main(String arg[])
{, c1 T' S% w0 ?; k7 ?8 X$ d* u
      int num []={1,2,3,4,5,6,7,8,9};0 a7 \8 Y* d: o; Y# f9 }, m
      while(trun)# V2 d( D, x  d6 s- K
{
int r=(int)(Math.random()*9)+1;
int t=num[i];! {, s: o- O1 x' F6 V1 m
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;  i# p- C4 M9 |6 p7 y
      int a2 = num[3] + num[4] + num[5] ;  X) S( G6 P8 V: m4 G  F
      int a3 = num[6] + num[7] + num[8] ;6 j+ K, W# I; P. r
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;) }9 W# t1 R; _: i% m
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15); ^* W5 f$ y. r+ ~7 K
      }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+ |6 V9 H  Z3 a( `) o8 N
{3 @7 V$ @5 q3 P5 V* }8 n( W
public static void main(String arg[])
      {4 p7 @! ~" A; M7 {9 y6 u6 W
         int num []={1,2,3,4,5,6,7,8,9};2 g6 w. E( [* W
      while(trun)
            {1 z5 u. a# L" N6 @( H! g9 b
            int r=(int)(Math.random()*9)+1;% L3 z- y3 }/ h( `% @* [' `- S
            int t=num[i];
            num[i]=num[r];
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;9 u3 a+ x' V6 x( u1 ^3 i
      int a2 = num[3] + num[4] + num[5] ;
      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] ;2 D" i& c8 @3 n5 l
      int a7 = num[0] + num[4] + num[8] ;7 E( h/ a0 h3 f  x+ C, H6 G; R, q) ^
      int a8 = num[2] + num[4] + num[6] ;& ?  e) t, E- l  _
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          % i/ @6 Z: ]7 T7 V3 \6 @
}7 }9 \- F6 ~+ m/ e
}

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