我正在阅读随机数和他们的一代.自从我开始编程以来,我对随机性很感兴趣.我读到Linux内核也使用随机数生成架构.
The structure consists of a two-level cascaded sequence of pools coupled with
CSPRNGs.
Each pool is a large group of bits which represents the current state of the
random number generator. The CSPRNGs are currently based on SHA-1, but the
kernel developers are considering a switch to SHA-3.
The kernel RNG produces two user-space output streams. One of these goes to
/dev/urandom and also to the kernel itself; the latter is useful because there
are uses for random numbers …我想执行一些计算,我希望结果正确到一些小数位,比如12.所以我写了一个样本:
#define PI 3.1415926535897932384626433832795028841971693993751
double d, k, h;
k = 999999/(2*PI);
h = 999999;
d = PI*k*k*h;
printf("%.12f\n", d);
但它给出了输出:
79577232813771760.000000000000
我甚至使用了setprecision(),但相同的答案却是指数形式.
cout<<setprecision(12)<<d<<endl;
版画
7.95772328138e+16
使用长双,但徒劳无功.
除了在long long int类型中分别存储整数部分和小数部分之外,还有什么办法吗？
如果是这样,可以做些什么来准确得到答案？
我只是想问一下这里发生了什么,我哪里出错了？
vector<int> a(5);
for(int i=0; i<5; i++) cin>>a[i]; //Input is 1 2 3 4 5
for(int i=0; i<5; i++) cout<<a[i]<<" "; //Prints correct, 1 2 3 4 5
cout<<endl;
for(VI::iterator it = a.begin(); it!=a.end(); ++it) {
cout<<a[*it]<<" "; //Prints incorrect output
}
cout<<endl;
看起来,错误输出中的最后一个元素是a[*(a.end()-1)]第一个元素,它实际上应该是缺少的.