洛谷P1000超级马里奥的神奇解法

　　话说上过洛谷的都知道，有一道经典例题P1000超级马里奥，这一题，可以说是非常简单非常经典，但是就算如此，还是可以人才辈出，我是个比较循规蹈矩的人（雾），所以我的代码就比较平常，也就是直接输出了所要求的超级马里奥的图片，那道题题目是这样的

超级玛丽是一个非常经典的游戏。请你用字符画的形式输出超级玛丽中的一个场景。

                ********
               ************
               ####....#.
             #..###.....##....
             ###.......######              ###            ###
                ...........               #...#          #...#
               ##*#######                 #.#.#          #.#.#
            ####*******######             #.#.#          #.#.#
           ...#***.****.*###....          #...#          #...#
           ....**********##.....           ###            ###
           ....****    *****....
             ####        ####
           ######        ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
##########################################    #----------#

#.....#......##.....#......##.....#......#    #----------#
##########################################    #----------#
#.#..#....#..##.#..#....#..##.#..#....#..#    #----------#
##########################################    ############
一般人都会写个cout+endl来实现输出的操作，但是，网友的力量是强大的，只有想不到没有做不到的，
接下来我介绍几种 ，注意，非原创

比如这位大佬，看之前先来一波膜拜orz，orz；

#include <bits/stdc++.h>
using namespace std;
int mp[100][100];
int last[100];
int n = 22, m = 62;
// 在[x1-x2, y1-y2]绘制ch
void draw(int x1, int y1, int x2, int y2, char ch = '#'){
    for(int i = x1; i <= x2; i++)
        for(int j = y1; j <= y2; j++)
            mp[i][j] = ch;
}
// 在[x1, y1]绘制ch
void draw(int x1, int y1, char ch = '#'){
    draw(x1, y1, x1, y1, ch);
}
// 以[x, y]为左上角绘制泥土
void drawland(int x, int y){
    draw(x, y, x+8, y+13);
    for(int i = x+1; i < x+8; i+=2)
        draw(i, y+1, i, y+12, '.');
    draw(x+1, y+4); draw(x+1, y+11);
    draw(x+3, y+3); draw(x+3, y+8);
    draw(x+5, y+6); draw(x+7, y+2);
    draw(x+7, y+5); draw(x+7, y+10);
}
// 以[x, y]为左上角绘制小岛
void drawisland(int x, int y){
    draw(x, y, x+3, y+19);
    draw(x+1, y+1, x+2, y+18, '-');
    draw(x+4, y+4, x+8, y+15);
    draw(x+4, y+5, x+7, y+14, '-');
}
// 以[x, y]为左上角绘制金币
void drawcoin(int x, int y){
    draw(x, y, x+5, y+4);
    draw(x+1, y+1, x+4, y+3, '.');
    draw(x+2, y+2, x+3, y+2);
    draw(x, y, ' '); draw(x+5, y, ' ');
    draw(x, y+4, ' '); draw(x+5, y+4, ' ');
}
// 以[x, y]为左上角绘制马里奥
void drawman(int x, int y){
    draw(x, y+5, x, y+12, '*'); x++;
    draw(x, y+4, x, y+15, '*'); x++;
    draw(x, y+4, x, y+7); draw(x, y+8, x, y+13, '.'); draw(x, y+12); x++;
    draw(x, y+2, x, y+14); draw(x, y+3, x, y+4, '.');
    draw(x, y+8, x, y+12, '.'); draw(x, y+15, x, y+18, '.'); x++;
    draw(x, y+2, x, y+17); draw(x, y+5, x, y+11, '.'); x++;
    draw(x, y+5, x, y+15, '.'); x++;
    draw(x, y+4, x, y+13); draw(x, y+6, '*'); x++;
    draw(x, y+1, x, y+17); draw(x, y+5, x, y+11, '*'); x++;
    draw(x, y, x+2, y+20, '.'); draw(x, y+4, x+2, y+16, '*');
    draw(x, y+3); draw(x, y+14, x+1, y+16); draw(x+1, y+16, '.');
    draw(x+2, y+8, x+2, y+11, ' '); draw(x, y+7, '.'); draw(x, y+12, '.');
    draw(x+3, y, x+4, y+19); draw(x+3, y+6, x+4, y+13, ' ');
    draw(x+3, y, x+3, y+1, ' '); draw(x+3, y+18, x+3, y+19, ' ');
}
// 打印输出
void printscreen(){
    for(int i = 1; i <= n; i++){
        last[i] = m;
        while(mp[i][last[i]] == ' ')
            last[i]--;
    }
    for(int i = 1; i <= n; i++,puts(""))
        for(int j = 1; j <= last[i]; j++)
            putchar(mp[i][j]);
}
int main(){
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            mp[i][j] = ' ';
    // 绘制人
    drawman(1, 12);
    // 绘制他脚下的三块泥土
    drawland(14, 1); drawland(14, 15); drawland(14, 29);
    // 绘制金币下面的那个岛屿
    drawisland(14, 43);
    // 绘制两个金币
    drawcoin(5, 43); drawcoin(5, 58);
    // 输出
    printscreen();
    return 0;
}

大致也看完了吧，这个人，不，这位大佬以坐标的方式，一个一个的把字符输出，最后达到了一样的效果，不得不说，这也太强了吧，你也去试试吧（雾）

如果只有这么点我也不会说是人才辈出了，下面这位可以说是暴力出奇迹了（雾）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<climits>
#include<cfloat>
#include<queue>
#include<cstddef>

using namespace std;
char x;
int a[10000]={32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,42,42,42,42,42,42,42,42,10,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,42,42,42,42,42,42,42,42,42,42,42,42,10,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,35,46,46,46,46,35,46,10,32,32,32,32,32,32,32,32,32,32,32,32,32,35,46,46,35,35,35,46,46,46,46,46,35,35,46,46,46,46,10,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,46,46,46,46,46,46,46,35,35,35,35,35,35,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,10,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,46,46,46,46,46,46,46,46,46,46,46,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,46,46,46,35,32,32,32,32,32,32,32,32,32,32,35,46,46,46,35,10,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,42,35,35,35,35,35,35,35,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,35,46,35,46,35,32,32,32,32,32,32,32,32,32,32,35,46,35,46,35,10,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,35,42,42,42,42,42,42,42,35,35,35,35,35,35,32,32,32,32,32,32,32,32,32,32,32,32,32,35,46,35,46,35,32,32,32,32,32,32,32,32,32,32,35,46,35,46,35,10,32,32,32,32,32,32,32,32,32,32,32,46,46,46,35,42,42,42,46,42,42,42,42,46,42,35,35,35,46,46,46,46,32,32,32,32,32,32,32,32,32,32,35,46,46,46,35,32,32,32,32,32,32,32,32,32,32,35,46,46,46,35,10,32,32,32,32,32,32,32,32,32,32,32,46,46,46,46,42,42,42,42,42,42,42,42,42,42,35,35,46,46,46,46,46,32,32,32,32,32,32,32,32,32,32,32,35,35,35,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,10,32,32,32,32,32,32,32,32,32,32,32,46,46,46,46,42,42,42,42,32,32,32,32,42,42,42,42,42,46,46,46,46,10,32,32,32,32,32,32,32,32,32,32,32,32,32,35,35,35,35,32,32,32,32,32,32,32,32,35,35,35,35,10,32,32,32,32,32,32,32,32,32,32,32,35,35,35,35,35,35,32,32,32,32,32,32,32,32,35,35,35,35,35,35,10,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,10,35,46,46,46,35,46,46,46,46,46,46,35,46,35,35,46,46,46,35,46,46,46,46,46,46,35,46,35,35,46,46,46,35,46,46,46,46,46,46,35,46,35,35,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,35,10,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,35,10,35,46,46,35,46,46,46,46,35,46,46,46,46,35,35,46,46,35,46,46,46,46,35,46,46,46,46,35,35,46,46,35,46,46,46,46,35,46,46,46,46,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,10,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,32,32,32,32,35,45,45,45,45,45,45,45,45,45,45,35,10,35,46,46,46,46,46,35,46,46,46,46,46,46,35,35,46,46,46,46,46,35,46,46,46,46,46,46,35,35,46,46,46,46,46,35,46,46,46,46,46,46,35,32,32,32,32,35,45,45,45,45,45,45,45,45,45,45,35,10,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,32,32,32,32,35,45,45,45,45,45,45,45,45,45,45,35,10,35,46,35,46,46,35,46,46,46,46,35,46,46,35,35,46,35,46,46,35,46,46,46,46,35,46,46,35,35,46,35,46,46,35,46,46,46,46,35,46,46,35,32,32,32,32,35,45,45,45,45,45,45,45,45,45,45,35,10,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,32,32,32,32,35,35,35,35,35,35,35,35,35,35,35,35,10};
int main()
{
    for(int i = 0;i <= 1148;i++)
    {
        cout<<(char)a[i]; 
    } 
    return 0;
}

打表。。。

这也未尝不是一种不错的方法，能存下所有的坐标，也是厉害所在啊%%%！这些还不够达到人才辈出，没错，还有优秀的代码优秀的人！

接下来是优秀的人，但看他的代码没什么出众的地方，但是他的题解可以说是很有文化了，我截取他的代码的其中一段的题解

吾有一術。名之曰「画点」。欲行是術。必先得三數。曰「甲」。曰「乙」。曰「字」。乃行是術曰。 施「画」於「甲」。於「乙」。於「甲」。於「乙」。於「字」。 是謂「画点」之術也。 吾有一術。名之曰「画地」。欲行是術。必先得二數。曰「甲」。曰「乙」。乃行是術曰。 加「甲」以八。名之曰「丙」。加「乙」以十三。名之曰「丁」。 施「画」於「甲」。於「乙」。於「丙」。於「丁」。於「井」。 吾有一列。名之曰「戊」。充「戊」以一。以三。以五。以七。 凡「戊」中之「己」。 加「己」以「甲」。名之曰「庚」。加「乙」以一。名之曰「丙」。加「乙」以十二。名之曰「丁」。 施「画」於「庚」。於「丙」。於「庚」。於「丁」。於「点」。 云云。

#include<stdio.h>
int main() {
    printf(
    "                ********
"
    "               ************
"
    "               ####....#.
"
    "             #..###.....##....
"
    "             ###.......######              ###            ###
"
    "                ...........               #...#          #...#
"
    "               ##*#######                 #.#.#          #.#.#
"
    "            ####*******######             #.#.#          #.#.#
"
    "           ...#***.****.*###....          #...#          #...#
"
    "           ....**********##.....           ###            ###
"
    "           ....****    *****....
"
    "             ####        ####
"
    "           ######        ######
"
    "##############################################################
"
    "#...#......#.##...#......#.##...#......#.##------------------#
"
    "###########################################------------------#
"
    "#..#....#....##..#....#....##..#....#....#####################
"
    "##########################################    #----------#
"
    "#.....#......##.....#......##.....#......#    #----------#
"
    "##########################################    #----------#
"
    "#.#..#....#..##.#..#....#..##.#..#....#..#    #----------#
"
    "########################

 

##################    ############
"
    );
    return 0;
}

看着题解，心里自愧不如，你说编程不比别人好也就算了，语文也被吊打。。。

今天的内容就只有这些了，在最后，请广大博主帮我看看，这个代码是什么意思？（雾）python学的少，没见过世面（大雾）

import zlib,base64;print zlib.decompress(base64.b64decode('eJylUkEOwCAIu/sKk95Mxv+fN2AsE0S3xR42sRYqUqtHM5TJfsaBQQyQZ3jTGLp+JcqIjBUMSUMYy1Z6EB1J0hCOrhuywnoeXo4gF5U1I0nxJhc3rKSmH2vDd/Ny+nkMa6I3F7xmallrgvRh0K9L4DzLPLZQ9IY2CdekTKNjwK/qqdwK4J7TeZSa/wxtWKjc3W4VjdKdqt2tsIp2q3Y4AbV3oO4=')).decode()