jva 104 河洛之數

1) 矩陣內數字皆不可重複出現2 v4 h2 M' W: U+ P) T5 t5 h' _
(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
{
public static void main(String arg[])
{
      int num []={1,2,3,4,5,6,7,8,9};( ^) Z1 e0 j0 H) i! d* K
      while(trun)
{
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];
num[r]=t;7 E# |3 r8 Q/ V  ?
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;& m/ P! s) i' |5 s+ b) C
      int a3 = num[6] + num[7] + num[8] ;* B  ~8 w3 B6 a) f- N/ l; J. k
      int a4 = num[0] + num[3] + num[6] ;; c& u# ?! i" P( a2 y% ]* ~
      int a5 = num[1] + num[4] + num[7] ;$ `  y! F& U4 H$ h' z
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;" b# T+ g$ n, g' e) N. ?
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)6 Z- ^# T  F- }  ]9 u# K
      }break;
}7 z* \. J& ]4 {) i: L" w* W# 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
{
public static void main(String arg[])
      {4 o- N+ B- K# l0 @' ^
         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];5 v, m/ V# `  ~# m0 A
            num[r]=t;( n# Q1 d% h8 }
            int a1 = num[0] + num[1] + num[2] ;8 O+ p; b. P0 j# R' a$ o
      int a2 = num[3] + num[4] + num[5] ;. F; t3 @- C' P6 W; g5 j
      int a3 = num[6] + num[7] + num[8] ;- V" v/ V1 G: w3 I
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;. E* |+ t) ]* \' B
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;3 @* k! w2 I# J, j1 ?) ]
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)$ O( _5 Q) w! B2 N% ^- j
         }break;          7 A$ @- D7 e% k% n9 d; A
}
}

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