我有以下几种方法。 rnd 方法返回两个边界之间的单个随机整数：
/* Create next batch of 55 random numbers */
void advance_random (){
int j1;
double new_random;
for(j1=0; j1<24; j1++){
new_random = oldrand[j1]-oldrand[j1+31];
if(new_random<0.0){
new_random = new_random+1.0;
}
oldrand[j1] = new_random;
}
for(j1=24; j1<55; j1++){
new_random = oldrand[j1]-oldrand[j1-24];
if(new_random<0.0){
new_random = new_random+1.0;
}
oldrand[j1] = new_random;
}
} //advance_ramdom
/* Fetch a single random number between 0.0 and 1.0 */
double randomperc(){
jrand++;
if(jrand>=55){
jrand = 1;
advance_random();
}
return((double)oldrand[jrand]);
} //randomPerc
/* Fetch a single random integer between low and high including the bounds */
synchronized int rnd (int low, int high){
int res;
if (low >= high){
res = low;
} else {
res = low + (int)(randomperc()*(high-low+1));
if (res > high){
res = high;
}
}
return (res);
} // rnd
如何修改它以使返回的数字 mod2 =0？
Thanks
如果你能得到一个范围内的随机数[a, b]那么你所要做的就是获得一个范围内的随机数[(a+1)/2, b/2]并将其乘以 2 以获得范围内的随机偶数[a, b]
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