永远的“超级玛丽游戏”（洛谷）
最近闲的蛋疼，来科普一道非常困难 简单的题目：超级玛丽游戏。不知道是哪个SB 大佬出的题目，打的我手都快废了，啊啊啊啊啊，下面，让我们来认真的学习一下这道题目。
先发个福利：超级玛丽
解法1：
#include
int main() {
printf(
" ********\n"
" ************\n"
" ####....#.\n"
" #..###.....##....\n"
" ###.......###### ### ###\n"
" ........... #...# #...#\n"
" ##*####### #.#.# #.#.#\n"
" ####*******###### #.#.# #.#.#\n"
" ...#***.****.*###.... #...# #...#\n"
" ....**********##..... ### ###\n"
" ....**** *****....\n"
" #### ####\n"
" ###### ######\n"
"##############################################################\n"
"#...#......#.##...#......#.##...#......#.##------------------#\n"
"###########################################------------------#\n"
"#..#....#....##..#....#....##..#....#....#####################\n"
"########################################## #----------#\n"
"#.....#......##.....#......##.....#......# #----------#\n"
"########################################## #----------#\n"
"#.#..#....#..##.#..#....#..##.#..#....#..# #----------#\n"
"########################################## ############\n"
);
return 0;
} //本蒟蒻手打的，虽然通过了，但是真的手快断了，cry......
下面来欣赏一下大佬们的优秀 鬼畜做法。
解法2：
#include
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;
}
解法3：
#include
int main()
{
std::cout< 解法4（C#）：
using System;
namespace Luogu
{
class Algo{
public static void Main()
{
Console.Write(@" ********
************
####....#.
#..###.....##....
###.......###### ### ###
........... #...# #...#
##*####### #.#.# #.#.#
####*******###### #.#.# #.#.#
...#***.****.*###.... #...# #...#
....**********##..... ### ###
....**** *****....
#### ####
###### ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
########################################## #----------#
#.....#......##.....#......##.....#......# #----------#
########################################## #----------#
#.#..#....#..##.#..#....#..##.#..#....#..# #----------#
########################################## ############");
}
}
}
解法5（python）：
print(""" ********
************
####....#.
#..###.....##....
###.......###### ### ###
........... #...# #...#
##*####### #.#.# #.#.#
####*******###### #.#.# #.#.#
...#***.****.*###.... #...# #...#
....**********##..... ### ###
....**** *****....
#### ####
###### ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
########################################## #----------#
#.....#......##.....#......##.....#......# #----------#
########################################## #----------#
#.#..#....#..##.#..#....#..##.#..#....#..# #----------#
########################################## ############""")
解法6（pascal）：
begin
writeln(' ********');
writeln(' ************');
writeln(' ####....#.');
writeln(' #..###.....##....');
writeln(' ###.......###### ### ###');
writeln(' ........... #...# #...#');
writeln(' ##*####### #.#.# #.#.#');
writeln(' ####*******###### #.#.# #.#.#');
writeln(' ...#***.****.*###.... #...# #...#');
writeln(' ....**********##..... ### ###');
writeln(' ....**** *****....');
writeln(' #### ####');
writeln(' ###### ######');
writeln('##############################################################');
writeln('#...#......#.##...#......#.##...#......#.##------------------#');
writeln('###########################################------------------#');
writeln('#..#....#....##..#....#....##..#....#....#####################');
writeln('########################################## #----------#');
writeln('#.....#......##.....#......##.....#......# #----------#');
writeln('########################################## #----------#');
writeln('#.#..#....#..##.#..#....#..##.#..#....#..# #----------#');
writeln('########################################## ############');
end.
解法7：
#include
int main()
{
printf(" ********\n");
printf(" ************\n");
printf(" ####....#.\n");
printf(" #..###.....##....\n");
printf(" ###.......###### ### ###\n");
printf(" ........... #...# #...#\n");
printf(" ##*####### #.#.# #.#.#\n");
printf(" ####*******###### #.#.# #.#.#\n");
printf(" ...#***.****.*###.... #...# #...#\n");
printf(" ....**********##..... ### ###\n");
printf(" ....**** *****....\n");
printf(" #### ####\n");
printf(" ###### ######\n");
printf("##############################################################\n");
printf("#...#......#.##...#......#.##...#......#.##------------------#\n");
printf("###########################################------------------#\n");
printf("#..#....#....##..#....#....##..#....#....#####################\n");
printf("########################################## #----------#\n");
printf("#.....#......##.....#......##.....#......# #----------#\n");
printf("########################################## #----------#\n");
printf("#.#..#....#..##.#..#....#..##.#..#....#..# #----------#\n");
printf("########################################## ############\n");
return 0;
}
解法8(PHP)：
********
************
####....#.
#..###.....##....
###.......###### ### ###
........... #...# #...#
##*####### #.#.# #.#.#
####*******###### #.#.# #.#.#
...#***.****.*###.... #...# #...#
....**********##..... ### ###
....**** *****....
#### ####
###### ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
########################################## #----------#
#.....#......##.....#......##.....#......# #----------#
########################################## #----------#
#.#..#....#..##.#..#....#..##.#..#....#..# #----------#
########################################## ############
解法9(python最短代码挑战赛)：
import zlib,base64;print zlib.decompress(base64.b64decode('eJylUkEOwCAIu/sKk95Mxv+fN2AsE0S3xR42sRYqUqtHM5TJfsaBQQyQZ3jTGLp+JcqIjBUMSUMYy1Z6EB1J0hCOrhuywnoeXo4gF5U1I0nxJhc3rKSmH2vDd/Ny+nkMa6I3F7xmallrgvRh0K9L4DzLPLZQ9IY2CdekTKNjwK/qqdwK4J7TeZSa/wxtWKjc3W4VjdKdqt2tsIp2q3Y4AbV3oO4=')).decode()
解法10(ascll版)：
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
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()
{
// 被我注释掉的部分是我当时拿来转换的程序
// freopen("x.in","r",stdin);
// freopen("x.out","w",stdout);
// x=getchar();
// while(x!=EOF)
// {
// printf("%d,",(int)x);
// x=getchar();
// }
for(int i = 0;i <= 1148;i++)
{
cout<<(char)a[i];
}
return 0;
}
解法11（pascal手打压缩）：
const
p:ansistring=' 9 3*6! 9 2*9*! 9 2#2.2#.! 9 0#.0#1.3#0.2! 9 0#1.5#4 9 1#1 9 #1! 9 3.9 9 2#.1# 8#.1#! 9 2#0*#5 9 4#.#.# 8#.#.#! 9 #2*5#4 9 0#.#.# 8#.#.#! 9.1#*1.*2.*#1.2 8#.1# 8#.1#! 9.2*8#0.3 9#1 9 #1! 9.2*2 2*3.2! 9 0#2 6#2! 9#4 6#4!#9#9#9#9#9#5!#.1#.4#.#0.1#.4#.#0.1#.4#.#0-9-5#!#9#9#9#8-9-5#!#.0#.2#.2#0.0#.2#.2#0.0#.2#.2#9#8!#9#9#9#7 2#-8#!#.3#.4#0.3#.4#0.3#.4# 2#-8#!#9#9#9#7 2#-8#!#.#.0#.2#.0#0.#.0#.2#.0#0.#.0#.2#.0# 2#-8#!#9#9#9#7 2#9#!';
var i,j:longint;
begin
for i:=1 to length(p) do
case p[i] of
'0'..'9':for j:=1 to ord(p[i])-ord('0')+1 do write(p[i-1]);
'!':writeln;
else write(p[i])
end
end.
解法12（Ruby）：
puts <好了，就这么多。