给定一个函数,它产生的是1到5之间的随机整数,写一个函数,它产生的是1到7之间的随机整数。
当前回答
extern int r5();
int r7() {
return ((r5() & 0x01) << 2 ) | ((r5() & 0x01) << 1 ) | (r5() & 0x01);
}
其他回答
extern int r5();
int r7() {
return ((r5() & 0x01) << 2 ) | ((r5() & 0x01) << 1 ) | (r5() & 0x01);
}
在php中
function rand1to7() {
do {
$output_value = 0;
for ($i = 0; $i < 28; $i++) {
$output_value += rand1to5();
}
while ($output_value != 140);
$output_value -= 12;
return floor($output_value / 16);
}
循环生成16到127之间的随机数,除以16生成1到7.9375之间的浮点数,然后舍入得到1到7之间的整数。如果我没记错的话,得到7个结果中的任何一个的概率都是16/112。
为什么不除以5再乘以7,然后四舍五入呢?(当然,你必须使用浮点数no.)
它比其他解决方案更简单、更可靠(真的吗?)例如,在Python中:
def ranndomNo7():
import random
rand5 = random.randint(4) # Produces range: [0, 4]
rand7 = int(rand5 / 5 * 7) # /5, *7, +0.5 and floor()
return rand7
这不是很容易吗?
Here is a solution that tries to minimize the number of calls to rand5() while keeping the implementation simple and efficient; in particular, it does not require arbitrary large integers unlike Adam Rosenfield’s second answer. It exploits the fact that 23/19 = 1.21052... is a good rational approximation to log(7)/log(5) = 1.20906..., thus we can generate 19 random elements of {1,...,7} out of 23 random elements of {1,...,5} by rejection sampling with only a small rejection probability. On average, the algorithm below takes about 1.266 calls to rand5() for each call to rand7(). If the distribution of rand5() is uniform, so is rand7().
uint_fast64_t pool;
int capacity = 0;
void new_batch (void)
{
uint_fast64_t r;
int i;
do {
r = 0;
for (i = 0; i < 23; i++)
r = 5 * r + (rand5() - 1);
} while (r >= 11398895185373143ULL); /* 7**19, a bit less than 5**23 */
pool = r;
capacity = 19;
}
int rand7 (void)
{
int r;
if (capacity == 0)
new_batch();
r = pool % 7;
pool /= 7;
capacity--;
return r + 1;
}
假设rand给予所有位相同的权重,然后用上界进行掩码。
int i = rand(5) ^ (rand(5) & 2);
Rand(5)只能返回:1b, 10b, 11b, 100b, 101b。有时候你只需要考虑设置2位。