jva 104 河洛之數

1) 矩陣內數字皆不可重複出現1 l3 C$ n5 H6 z+ P! O, H, J
(2) 程式每次執行，矩陣數字排列皆不相同" K- n: x' R9 W; e. ]% }2 N! s, T" u' k
(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;
}

複製代碼

新手上路

Rank: 1

2^#

johnson發表於 2011-12-31 11:35 | 只看該作者

public class j1021 K; ~" A" T9 R2 ]& P- ~+ j
{. ]# l4 s5 U6 k0 g2 y8 c, A
public static void main(String arg[])1 I4 _# d" b6 g# Y9 N& G" p
{
      int num []={1,2,3,4,5,6,7,8,9};8 p2 m0 l- B& }7 C- j" N
      while(trun)3 K: y  m- \2 M6 i: O
{
int r=(int)(Math.random()*9)+1;
int t=num[i];5 Q9 @+ P- M" P! S8 W  N/ ^
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;3 L* X% Q( q/ T2 s
      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] ;0 D0 N( l6 [: _* M4 D+ E
      int a7 = num[0] + num[4] + num[8] ;: F, k" d% `) Q  I
      int a8 = num[2] + num[4] + num[6] ;6 W6 c' G* P2 Y
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;       ! d' P; R. J' ~0 d) ~
}2 I" L5 Y' Y" }( R5 ~
}

小雲雀

TOP

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;
}

複製代碼

★ 嘉凱~~☆

TOP

中級會員

Rank: 3 Rank: 3

4^#

eric5802發表於 2011-12-31 11:38 | 只看該作者

public class j102
{
public static void main(String arg[])
      {6 I, B* f1 [, s" B% G. H8 w% `
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)5 N; D9 W5 o% t% s
            {" z+ h* _* x6 E1 }% N* n+ T
            int r=(int)(Math.random()*9)+1;
            int t=num[i];7 y* U2 C' e. _' P$ L4 S* P
            num[i]=num[r];
            num[r]=t;/ u2 o, i% ]3 k
            int a1 = num[0] + num[1] + num[2] ;2 h& O1 G$ e! U  ?
      int a2 = num[3] + num[4] + num[5] ;1 F/ z9 n- r$ H: E
      int a3 = num[6] + num[7] + num[8] ;+ G- p# x" w3 c7 U. w# |
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;% j, K- H, @9 i7 ^3 C
      int a6 = num[2] + num[5] + num[8] ;/ T# k1 P2 ]( Y
      int a7 = num[0] + num[4] + num[8] ;1 k1 M* f1 b- |! N' W/ Z
      int a8 = num[2] + num[4] + num[6] ;7 K9 {9 N8 u+ |& D6 V8 P
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)& z) P# d4 J4 O+ i' N3 `9 R
         }break;          ! B0 w* Z. Z- n7 c- B
}
}

人平

TOP