c/c++随机数
问个问题 for (i=0;i <6;i++)
{
srand( (unsigned)time( NULL )+i);
tmpnum = rand()%10;
for(j=i;j <5;j++)
tmpnum = tmpnum*10;
num = num + tmpnum;
}
如果2次调用的时间很短.就会产生一样的的结果了 就是最后得到的num值都是一样额
怎么避免

[解决办法]
srand( (unsigned)time( NULL ))放循环外面就可以了, 没有必要每次都改变种子吧?

[解决办法]
#include "iostream.h "
#include "time.h "
#include "stdlib.h "

int main(int argc, char* argv[])
{
time_t t;
srand((unsigned)time(&t));
cout < < "Ten random numbers from 0 -- 99 " < <endl;;
for(int i = 0; i < 10; i++)
{
cout < <rand()%100 < < " ";
}
cout < <endl;
return 0;
}
[解决办法]
srand( (unsigned)time( NULL )+i); //种种子的过程放到外面
for (i=0;i <6;i++)
{
tmpnum = rand()%10;
for(j=i;j <5;j++)
tmpnum = tmpnum*10;
num = num + tmpnum;
}
[解决办法]
#include "iostream.h "
#include "time.h "
#include "stdlib.h "

int main() //主函数没必要加参数吧？
{
time_t t;
srand((unsigned)time(&t));
cout < < "Ten random numbers from 0 -- 99 " < <endl;;
for(int i = 0; i < 10; i++)
{
cout < <rand()%100 < < " ";
}
cout < <endl;
return 0;
}