分布式 反熵

来源:互联网 发布:如何破解MD5的网络包 编辑:程序博客网 时间:2024/06/13 01:11

介绍:n个离散值,经过数次循环,各个离散值无限接近于n值的平均值

#include <iostream>#include <vector>#include <string>#include <stdlib.h> #include <time.h>  using namespace std;int main() {    srand((unsigned)time(NULL));    int n,t;    double sum=0;    n = 10 + rand() % 10;    t = 10 + rand() % 10;    cout << "总共" << n << "个点" <<"循环"<<t<<"次"<< endl;    vector<double> v(n);    cout << "初始值:" << endl;    for (int i = 0; i < n; i++) {        double temp = 1 + rand() % 99;        v[i] = temp;        sum += temp;        cout << temp << " ";    }    cout << endl;    for (int i = 0; i <t; i++) {        for (int j = 0; j < n; j++) {            int jj = 0 + rand() % n;            v[j] = (v[j] + v[jj]) / 2;            v[jj] = v[j];        }        cout << "第" << i + 1 << "次循环后:" << endl;        for (int j = 0; j < n; j++)        {            cout << v[j] << " ";        }        cout << endl;    }    cout << endl;    cout << "预设平均值:" << sum / n << endl;    system("pause");    return 0;}
原创粉丝点击