产生100个随机下落的点(标签-随机数|关键词-随机、动画)

Question

想要在屏幕上随机产生100个下落的点

#include <graphics.h>
#include <conio.h>
#include <iostream>
using namespace std;
#pragma warning(disable:4996)

void Point(int x, int y)
{
    for (int j = 0; j < 600; j++)
    {
        setrop2(R2_XORPEN);
        putpixel(x, y + j, WHITE);
        Sleep(5);
        putpixel(x, y + j, WHITE);
    }
}

int main()
{
    initgraph(600, 600);
    srand(time(0));
    int x, y = 0;
    int j = 0;
   


   for(int i=0;i<100;i++)
    {
       x = rand() % 599;
        y = rand() % 599;
       Point(x, y);
     
    }
   
        

    
   
   
    _getch();
    closegraph();
    return 0;
}

只出现一个随机下落的点，与题目要求不符了

我的解答思路和尝试过的方法

想要产生100个随机的点

Answer 1

这么改下，供参考：

#include <stdio.h>
#include <graphics.h>
#include <conio.h>
#include <stdlib.h>
#include <time.h>
//using namespace std;
#pragma warning(disable:4996)

void Point(int x, int y)
{
    for (int j = 0; j < 600 - y; j++) //for (int j = 0; j < 600; j++)
    {
        setrop2(R2_XORPEN);
        putpixel(x, y + j, WHITE);
        Sleep(4);
        putpixel(x, y + j, WHITE);
    }
}

int main()
{
    initgraph(600, 600);
    srand((unsigned int)time(NULL));//srand(time(0));
    int x, y = 0;
    int j = 0;
    for (int i = 0; i < 100; i++)
    {
        x = rand() % 599;
        y = rand() % 599;
        Point(x, y);
    }
    _getch();
    closegraph();
    return 0;
}