jva 104 河洛之數

1) 矩陣內數字皆不可重複出現/ L/ {: ~; c. A* {; a$ c* y
(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
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public static void main(String arg[])6 K! u8 a. g4 O
{2 Q2 b+ G" K& e0 \5 k* R; b
      int num []={1,2,3,4,5,6,7,8,9};2 R2 q3 r+ s7 @3 ^+ q# m
      while(trun)8 k9 l1 K. Q' ?% ?8 _' R7 |) I7 E7 a- j
{8 f6 p$ u! H9 n! N! [& N
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];
num[r]=t;+ q" ^4 N- C2 b& i" z
int a1 = num[0] + num[1] + num[2] ;' I$ G3 M- j2 E7 ?8 {" a. J; \
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;% P; {2 P. x* `( }( 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] ;
      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)& g; [  s" o* \, _; F: c  M
      }break;         |9 ?) U- x0 F( L1 k5 a$ d* O) |* G
}3 g; P- C  e3 {5 h( t
}

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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+ @3 U5 n3 \0 [1 l7 p( i
{
public static void main(String arg[]), Z+ P/ S6 ?7 L5 C; a, |
      {
         int num []={1,2,3,4,5,6,7,8,9};6 h, `4 N8 j, ~1 ~! m
      while(trun)
            {7 h8 d6 H/ e! r9 `
            int r=(int)(Math.random()*9)+1;
            int t=num[i];
            num[i]=num[r];3 L* }/ w# B- K  y% q
            num[r]=t;& ^1 N: V" m% V4 K; f. n
            int a1 = num[0] + num[1] + num[2] ;1 a' z0 m9 J, W: }' J
      int a2 = num[3] + num[4] + num[5] ;* K5 k$ i; O3 E! R- K3 a: i
      int a3 = num[6] + num[7] + num[8] ;. @% \, o6 d, |0 ^' Y8 k: e" F
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;9 G4 X5 z: ~4 i& m" x1 ]
      int a6 = num[2] + num[5] + num[8] ;# U7 D% U# r4 P6 }7 u& M. K, m# _2 I
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;4 C* {/ o- Q% v0 W* Y# W% q3 ]3 n7 q
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          2 O" Y7 h# K4 S2 Z; t
}! [6 G0 T: h6 c  L" M9 s; l
}

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