需要生成从100到500的随机数

Question

对于我的班级任务，我必须生成从100-500范围内的120个随机数。这是我现在使用的代码。
下面是输出的剪切和粘贴：我期望数字范围在100到500之间。我不确定我做错了什么。我没有得到理想的结果。我如何解决它以获得100-500的数字？

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void create_random_numbers(int ran[], int x) 
{ 
    unsigned seed = time(0); 

    srand(seed); 
    int random_integer; 
    for (int index = 0; index<400; index++) 
    { 

    // cout << random_integer << endl<<endl; 
     ran[index] = (rand()%500)+100; 

     cout << "ran "<< ran[index] << endl << endl; 
    } 

} 

Answer 1

rand() % 500产生的值在包括[ 0...499 ]的范围内，因此100 + [ 0 ... 499 ]产生值[ 100... 599 ]（含）。您需要rand() % 400 + 100。

看看下面的校正代码：

void create_random_numbers(int ran[]) 
{ 
    unsigned seed = time(0); 

    srand(seed); 
    int random_integer; 
    for (int index = 0; index < 120; index++) 
    { 
     ran[index] = (rand() % 400) + 100; 
     cout << index << " : " << ran[index] << endl << endl; 
    } 
} 

查看我已演示：Demo

Answer 2

另一种方法是使用均匀INT分布 产生了随机的整数值I，在均匀地分布闭区间[a，b]，即根据离散概率分布

function 
P(i|a,b) = 1/(b − a + 1) 
. 
#include <random> 
#include <iostream> 
#include <array> 

int main() 
{ 
    std::array<int, 120> numbers; 
    std::random_device rd; //Will be used to obtain a seed for the 
          //random number engine 
    std::mt19937 gen(rd()); //Standard mersenne_twister_engine seeded 
          // with rd() 
    std::uniform_int_distribution<> dis(100, 500); 

    for (int n = 0; n < 120; ++n) 
     //Use dis to transform the random unsigned int generated by gen 
     //into an int in [100, 500] 
     numbers[n] = dis(gen); 

    for (auto i : numbers) { 
     std::cout << ' ' << i; 
    } 
    std::cout << '\n'; 
    return 0; 
}