jva 104 河洛之數

1) 矩陣內數字皆不可重複出現% r# {) u5 [3 Y' n5 W1 j
(2) 程式每次執行，矩陣數字排列皆不相同" a7 P7 \! d5 n9 h: q" E
(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[])# ~# U% X; E5 J4 A  O4 C
{! S+ l+ a! G  N+ q. K, t3 b( O
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];6 u( u7 M- ~( L0 G6 _
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;+ q% n0 {) c  U, L2 t) n" R
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;6 ]) w) ?9 o/ N3 f. m/ K" W
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;9 v7 U# y* ^! J0 E
      int a8 = num[2] + num[4] + num[6] ;; l( s' F  y# [5 ^" e2 C, \  G
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)& f) `- C) V; f0 _
      }break;
}4 ]2 N0 j6 o: Y) L
}

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新手上路

Rank: 1

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$ D" t0 I! H( f  a1 `) k& E. t
{
public static void main(String arg[])
      {- f* D, d4 {' h0 K# W5 _$ b
         int num []={1,2,3,4,5,6,7,8,9};' c( Z5 t* x; s/ e
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;
            int t=num[i];' x: n# N* w* O* F! I
            num[i]=num[r];
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;( I  o: i8 R8 w+ P& S  l
      int a3 = num[6] + num[7] + num[8] ;8 f- Q) S0 e: }; h
      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] ;# t$ V4 M8 ?2 z2 `) |
      int a8 = num[2] + num[4] + num[6] ;: d& Y' L4 a8 u" ~, f( X# d+ ^
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          5 ?: }9 T& R6 q; y5 j
}
}

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