實際效果呈現彩色的動態閃動,實作效果也很簡單,只是用到了基本的回圈
一些函式查一查就好了,后續我可能會加上注釋(也沒什么好加的)
這個創意來自我的同學 s y q syq syq,這是他的博客博客鏈接
雖然程式短短幾十行不難,但是能有這個創意用簡單的回圈寫出漂亮的圖形
我覺的也很厲害了!!!原始碼在下面
#include <bits/stdc++.h>
#include <windows.h>
#include <conio.h>
using namespace std;
int len;
struct Dot {
int x;
int y;
char c;
}a[10000000];
int vis[10000000];
char c[5] = {35, 36, 37, 38, 79};
string s = "/\\";
int n;
string me = "Merry Christmas ^_^ !";
string fm = "From Yukisong";
void getcur(short int x,short int y)//移動游標函式
{
COORD pos = (COORD){x,y};//定義一個螢屏上的座標
HANDLE handle = GetStdHandle( STD_OUTPUT_HANDLE );
SetConsoleCursorPosition( handle,pos );;
}
void goprint(int x,int y, char c) {
SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), rand()%16);
getcur(x,y); printf("%c", c); //getcur(0,row);//消除游標影響
}
void del(int x,int y)
{
getcur(x,y); printf(" ");
}
int main() {
cout << "請輸入你喜歡的一個數字（10 - 17），并將視窗最大化：";
cin >> n;
cout << endl;
int Layer = 2 * n - 1;
int Space = 2 * n - 2;
for (int i = 1; i <= Layer; ++i) {
if (i & 1) {
for (int j = 0; j < 2 * i; j += 2) {
a[len].x = j + Space + 30, a[len].y = i + 5, a[len++].c = c[rand()%5];
}
Space -= 2;
}
else {
int r = 0;
for (int j = 0; j < 2 * i; j += 2) {
a[len].x = j + Space + 31, a[len].y = i + 5, a[len++].c = s[r];
r = (r + 1) % 2;
}
}
}
for (int i = 0; i < Layer / 3 + 1; ++i) {
a[len].x = Layer + 29, a[len].y = i + 6 + Layer, a[len++].c = '|';
a[len].x = Layer + 30, a[len].y = i + 6 + Layer, a[len++].c = '|';
}
for (int i = 0; i < me.length(); ++i) {
a[len].x = Layer - 1 + i + 30;
a[len].y = Layer / 3 + 10 + Layer;
a[len++].c = me[i];
}
for (int i = 0; i < fm.length(); ++i) {
a[len].x = Layer - 1 + i + 30;
a[len].y = Layer / 3 + 11 + Layer;
a[len++].c = fm[i];
}
while(1) {
int i = rand()%len;
if ((a[i].c == '\\' || a[i].c == '/') && vis[i]) {
continue;
}
goprint(a[i].x, a[i].y , a[i].c);
vis[i] = 1;
}
}
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