jva 104 河洛之數

1) 矩陣內數字皆不可重複出現
(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 j1020 A2 x0 s  m, I6 ^2 h" |, i, I
{
public static void main(String arg[])- W. {/ R0 i0 @7 x
{
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)5 b+ t: L# h# f" U' }
{
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];) r+ a/ `( m( j5 Q/ T2 [
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;. \4 q9 ?& F( j# v( a- X# o
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;5 S, C; z/ N( f. {9 G" ]
      int a6 = num[2] + num[5] + num[8] ;% I$ M2 _  E6 J
      int a7 = num[0] + num[4] + num[8] ;% n$ `# S' Y% [4 F
      int a8 = num[2] + num[4] + num[6] ;+ |4 Y( }# B$ q8 h
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)) Z( r0 K$ s, x/ {' l
      }break;
}
}

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新手上路

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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( Y& Q1 G; F" r: B, t* k6 Q
{% s4 V1 N& Y  X& z' r
public static void main(String arg[])
      {' D2 v. V# V; c4 W* d' t4 d
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;  c5 d0 G& U$ I; \2 P' _
            int t=num[i];" h" z* V' L  x& d9 i6 b- k& f; z- z
            num[i]=num[r];4 T5 [. [1 Y  x2 e3 ?
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;$ c! {7 x8 D2 s  R
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;+ Y. x/ c' s: Y8 p& H2 {3 G) ^6 ?+ Y
      int a6 = num[2] + num[5] + num[8] ;; f* _6 w- C. D, X
      int a7 = num[0] + num[4] + num[8] ;% w7 S  C* J+ D+ @7 A$ j: c
      int a8 = num[2] + num[4] + num[6] ;2 s8 g" P! m$ [! j
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)' `0 T1 M3 r! x6 M* A
         }break;
}
}

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