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Home Backend Development PHP Tutorial 3个数的和在一个范围里,求3个数

3个数的和在一个范围里,求3个数

Jun 23, 2016 pm 01:36 PM

这3个数的取值范围是1-6,必定是整数

第一种情况,3个数的和取值范围4-10,随机产生这3个数(分3个数可以全部重复和不可以全部重复两种情况)

第二种情况,3个数的和取值范围11-17,随机产生这3个数(分3个数可以全部重复和不可以全部重复两种情况)


 举例说明:
 如果允许全部重复,那么如果3个数的和是15,那么3个数可能是5,5,5,而如果不允许全部重复,
3个数只可能是4,5,6或,6,6,3或3,6,6之类的组合

 求一个函数,

//根据给出的和,和是否允许3个数完全相同3个数的数组,
//要求必须能否返回所有可能的组合,比如6,6,3和3,6,6都必须有可能返回。
//程序只要求随机返回所有可能性组合之一即可。
function get3nums($第几种情况,$是否允许全部重复)
//比如list($rand1,$rand2,$rand3) = get3nums(2,false);//第二种情况,且3个数字不允许全部相同
//得到$rand1=6,$rand2=6,$rand3=3

 function get3nums($qingkuang,$ifxiangtong){

 }


回复讨论(解决方案)

第一种情况
因为有三个数,可以先在传入数值的基础上减去2,然后随机出第一个数,然后计算剩余。
比如
function f($n)
{
$t = $n - 2;
$a = rand(1,min(6,$t));
$t = $n - $a - 1;
$b = rand(1,min(6,$t));
$t = $n - $a - $b;
return array($a,$b,$c);
}
如果要不能三个重复,增加判断重复的语句好了

上面有一个变量名写错了
function f($n)
{
$t = $n - 2;
$a = rand(1,min(6,$t));
$t = $n - $a - 1;
$b = rand(1,min(6,$t));
$c = $n - $a - $b;
return array($a,$b,$c);
}

第二种情况和第一种一样
不能全部重复只要在返回值前增加一个判断
if($a == $b && $b == $c)
{
return f($n);
}

/** * @$start  范围下限 * @$end    范围上限 * @$repeat 允许重复,默认不允许 **/ function get3nums($start, $end, $repeat=false) {  do {    $r = array();    $n = rand($start, $end); //取得指定范围中的一个数    $a = intval($n/3);    $r[] = rand(1, min($a, 6));    $r[] = rand(1, min($a, 6));    $r[] = $n - array_sum($r);    if($repeat) break;  }while(count(array_count_values($r)) == 1);  return $r;}print_r(get3nums(4, 10));print_r(get3nums(4, 10, 1));print_r(get3nums(11, 17));print_r(get3nums(11, 17, 1));
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/** * @$start  范围下限 * @$end    范围上限 * @$repeat 允许重复,默认不允许 **/ function get3nums($start, $end, $repeat=false) {  do {    $r = array();    $n = rand($start, $end); //取得指定范围中的一个数    $a = intval($n/3);    $r[] = rand(1, min($a, 6));    $r[] = rand(1, min($a, 6));    $r[] = $n - array_sum($r);    if($repeat) break;  }while(count(array_count_values($r)) == 1);  return $r;}print_r(get3nums(4, 10));print_r(get3nums(4, 10, 1));print_r(get3nums(11, 17));print_r(get3nums(11, 17, 1));
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不对啊,我的数字1-6之间,你这个好多都超过6了

Array
(
[0] => 2
[1] => 2
[2] => 4
)
Array
(
[0] => 1
[1] => 1
[2] => 2
)
Array
(
[0] => 4
[1] => 1
[2] => 8
)
Array
(
[0] => 2
[1] => 1
[2] => 8
)

不对啊,我的数字1-6之间,你这个好多都超过6了


你到底有没有看懂?


不对啊,我的数字1-6之间,你这个好多都超过6了


你到底有没有看懂?



3个数的取值都必须在1-6之间。输出的居然有8

/**
* @$start 范围下限
* @$end 范围上限
* @$repeat 允许重复,默认不允许
**/
function get3nums($start, $end, $repeat=false) {
if(!$repeat){
$n =rand($start, $end); //取得指定范围中的一个数
for($i=1;$i<7;$i++){
for($j=1;$j<7;$j++){
$k=$n-$i-$j;
if($k<7 &&$k>0){
if(($i==$j and $j==$k)){
continue;
}
echo $i.'-'.$j.'-'.$k."
";
}
}
}
}else{
$n =rand($start, $end);
$avg=$n/3;
if( $avg!=intval($avg)){
echo "no value"."
";
}else{
echo $avg.'-'.$avg.'-'.$avg."
";
}
}
}

print_r(get3nums(4, 10));
echo '
';
print_r(get3nums(4, 10, 1));
echo '
';
print_r(get3nums(11, 17));
echo '
';
print_r(get3nums(11, 17, 1));
?>

再检查一下就是了

function get3nums($start, $end, $repeat=false) {  do {    do {      $r = array();      $n = rand($start, $end); //取得指定范围中的一个数      $a = intval($n/3);      $r[] = rand(1, min($a, 6));      $r[] = rand(1, min($a, 6));      $r[] = $n - array_sum($r);    }while(max($r) > 6);    if($repeat) break;  }while(count(array_count_values($r)) == 1);  return $r;}
Copy after login

    /**
 * @$start  范围下限
 * @$end    范围上限
 * @$repeat 允许重复,默认不允许
 **/ 
function get3nums($start, $end, $repeat=false) {
  if(!$repeat){
      $n =rand($start, $end); //取得指定范围中的一个数
      for($i=1;$i           for($j=1;$j               $k=$n-$i-$j;
              if($k0){
                  if(($i==$j and $j==$k)){
                      continue;
                  }
                  echo $i.'-'.$j.'-'.$k."
";
              }
          }
      }
  }else{
      $n =rand($start, $end);
      $avg=$n/3;
      if( $avg!=intval($avg)){
          echo "no value"."
";
      }else{
          echo $avg.'-'.$avg.'-'.$avg."
";
      }
  }
}
 
print_r(get3nums(4, 10));
echo '
';
print_r(get3nums(4, 10, 1));
echo '
';
print_r(get3nums(11, 17));
echo '
';
print_r(get3nums(11, 17, 1));
?>



错误还是比较多的。而且11,17这种铁定出现6,不理想。
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