C语言实现520表白代码祝你表白成功!

Z时代
2024-01-10
分类：IT

本文实例为大家分享了C语言实现520表白的具体代码，供大家参考，具体内容如下

在这个狂虐单身狗的节日，实在无聊写了几行也不知道向谁告白的代码，献丑献丑：


#include<stdio.h>
#include<math.h>
int main()
{
 float y, x, z;
 printf("那一天\n");
 printf("第一次遇见你\n");
 printf("忘不了\n");
 printf("你的容颜\n");
 printf("若轻云之蔽月,如流风之回雪\n");
 printf("\n\n\n");
 printf("其实\n");
 printf("有一句话\n");
 printf("我一直想对你说:\n");
 for (double y = 2.5; y >= -1.6; y = y - 0.2)
 {
 for (double x = -3; x <= 4.8; x = x + 0.1)
 {
  (pow((x*x + y*y - 1), 3) <= 3.6*x*x*y*y*y
  || (x>-2.4 && x<-2.1 && y<1.5 && y>-1)
  || (((x<2.5 && x>2.2) || (x>3.4 && x<3.7)) && y>-1 && y<1.5)
  || (y>-1 && y<-0.6 && x<3.7 && x>2.2)) ? printf("*") : printf(" ");
 }
 printf("\n");
 }
 getchar();
}

运行效果图：

代码二


#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define I 20
#define R 340
#include <string.h>
int main()
{
	char answer[10];
	printf("遇到你\n我才发现\n曾经所有的条件\n似乎都成了我等你的借口\n\n");
	printf("请允许我，从今往后映入你\n明媚的眼\n");
	printf("输入yes或no,点击Enter");
	printf("\n  ");
	scanf("%s", answer);
	char b[4]="yes";
	if(strcmp(answer,b)==0)
    {
        float y, x, z, f;
	for (y = 1.5f; y > -1.5f; y -= 0.1f)
	{
		for (x = -1.5f; x < 1.5f; x += 0.05f)
		{
			z = x * x + y * y - 1;
			f = z * z*z - x * x*y*y*y;
			putchar(f <= 0.0f ? "*********"[(int)(f*-8.0f)] : ' ');
		}
		putchar('\n');
	}
	long time;
	for (;;)
	{
		system("color a");
		for (time = 0; time<99999999; time++);
		system("color b");
		for (time = 0; time<99999999; time++);
		system("color c");
		for (time = 0; time<99999999; time++);
		system("color d");
		for (time = 0; time<99999999; time++);
		system("color e");
		for (time = 0; time<99999999; time++);
		system("color f");
		for (time = 0; time<99999999; time++);
		system("color 0");
		for (time = 0; time<99999999; time++);
		system("color 1");
		for (time = 0; time<99999999; time++);
		system("color 2");
		for (time = 0; time<99999999; time++);
		system("color 3");
		for (time = 0; time<99999999; time++);
		system("color 4");
		for (time = 0; time<99999999; time++);
		system("color 5");
		for (time = 0; time<99999999; time++);
		system("color 6");
		for (time = 0; time<99999999; time++);
		system("color 7");
		for (time = 0; time<99999999; time++);
		system("color 8");
		for (time = 0; time<99999999; time++);
		system("color 9");
	}
	getchar();
    }
    else
    {
        printf("\n好吧，那就相忘于江湖吧！\n");
    }
    return 0;
}