纪念我逝去的astar, 和未提交的代码

doyouloveme posted @ 2012年6月17日 02:42 in astar with tags astar , 1179 阅读
……转载随便你注明不注明来源……只要你不说是你写的就行了……

A*2012,是如此的坑爹,当然我很弱是一个原因。

复赛没交一题……我还能说什么呢……考挂自己弱吧……比赛结束后15min,把round2b C的样例调出来了

round2b啊2b啊2b啊……

 

/*
* Copyright (C) 2012 All rights reserved.
* File name:  C.cpp
* Author:     Parabola_ds
* Creat time: 2012.06.17
*/
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;

const int inf = ~0U >> 1;

set<int> n;
char bd[600];
int maxl, len, space, t;
long long num;
bool tt[600];

void ps(){ rep(i, space) printf(" "); }

int S[600];

void solve(){
	space = 4;
	int m = 0, fst = 0;
	int num;
	for(int i = 0; i <= maxl;){
		int j, tmp;
		bool bl;
		m++;
		switch(bd[i]){
			case '3':
				fst++;
				ps(); printf("for (int r%d = 0; r%d < ", fst - 1, fst - 1);
				num = 0;
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				printf("%d; r%d++)", num, fst - 1);
				if(tt[m]) printf(" {");
				puts("");
				space += 4;
				i = j + 1;
				break;
			case '2':
				ps(); printf("if (");
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				if(bl) printf("r%d ", num); else printf("%d ", num);
				i = j + 1;
				if(bd[i] == '1') printf("== "); else if(bd[i] == '2') printf("< "); else printf("> ");
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				if(bl) printf("r%d ", num); else printf("%d ", num);
				i = j + 1;
				break;
			case '1':
				ps();
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				if(bl) printf("r%d =", num);
				i = j + 1;
				num = 0; bl = bd[i] == '#';
				j = i + (bd[i] == '#' ? 1 : 0);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				if(bl) printf(" r%d ", num); else printf(" %d ", num);
				i = j + 1;
				if(bd[i] == '1') printf("+"); else if(bd[i] == '2') printf("-"); else if(bd[i] == '3') printf("*"); else printf("/");
				tmp = i;
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				for(; bd[j] >= '0' && bd[j] <= '9' && j <= maxl; ++j)
					num = num * 10 + bd[j] - '0';
				if(bl) printf(" r%d;\n", num); else printf(" %d;\n", num);
				i = j + 1;
				break;
			case '0':
				space -= 4;
				i++;
				fst--;
				if(tt[m]) ps(), printf("}\n");
				break;
		}
	}
}

int main(){
	cin.getline(bd, 600);
	len = strlen(bd);
	int m = 0, fst = 0;
	for(int i = 0; i < len;){
		int j, tmp;
		bool bl;
		m++;
		switch(bd[i]){
			case '3':
				S[++fst] = m;
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				break;
			case '2':
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				if(bd[i] < '1' || bd[i] > '3')
					goto st;
				maxl = i;
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				break;
			case '1':
				num = 0; bl = bd[i + 1] == '#';
				if(!bl){
					maxl = i;
					goto st;
				}
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				num = 0; bl = bd[i] == '#';
				j = i + (bd[i] == '#' ? 1 : 0);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				if(bd[i] < '1' || bd[i] > '4')
					goto st;
				tmp = i;
				num = 0; bl = bd[i + 1] == '#';
				j = i + (bd[i + 1] == '#' ? 2 : 1);
				if(j >= len){
					maxl = i; goto st;
				}
				maxl = j;
				for(; bd[j] >= '0' && bd[j] <= '9'; ++j){
					num = num * 10 + bd[j] - '0';
					if(num < inf) maxl = i;
					if(num < inf) maxl = i; else{
						if(bl) n.insert(num);
						goto st;
					}
				}
				if(bl) n.insert(num);
				if(bd[tmp] == 4 && !num){
					maxl = tmp;
					goto st;
				}
				if(bd[j] != '*')
					goto st;
				maxl = j;
				i = j + 1;
				break;
			case '0':
				maxl = i++;
				if(m - S[fst] > 2) tt[S[fst]] = tt[m] = 1;
				fst--;
				break;
		}
	}
st:{}
	printf("%d\n", maxl + 1);
	printf("int main() {\n");
	for(set<int>::iterator it = n.begin(); it != n.end(); ++it)
		printf("    int r%d = 0;\n", *it);
	solve();
	printf("    printf(\"");
	set<int>::iterator it = n.begin();
	if(it != n.end()) printf("%%d"); it++;
	for(; it != n.end(); it++)
		printf(" %%d");
	printf("\"");
	for(it = n.begin(); it != n.end(); it++)
		printf(", r%d", *it);
	printf(");\n    return 0;\n}\n");
	return 0;
}

附题目(

    这题是用来防Ak的我一开始就跑去写干嘛啊!!!!!!!!!!!!!!!!!!!!!!!

    有点若菜的自我修养就不行么!!!!!!!!!!!!!!!!!!!!!

):

BD语言翻译器

 

  • 时间限制: 
  • 1000ms
  • 内存限制: 
  • 262144kB
  • 描述
  • 度度熊每天上下班都要被堵在路上至少两个小时。在这段时间里,他只能靠玩手机游戏打发时间。

    这天,度度熊又被堵在路上了。望着自己的手机,度度熊突发奇想:如果能够用自己的手机来写程序,不就可以充分利用这些宝贵的时间了么?

    当然,用手机写程序最大的坏处在于,手机只有12个键:10个数字键、一个“#”键和一个“*”键。因此,度度熊特意设计了一种只包含有这些符号的语言——Best Digital (简称BD语言)。这样的话,度度熊就可以在路上用手机编写BD程序,然后使用翻译器将其直接翻译为对应的C++程序。

    虽然度度熊一下就知道怎么实现这个翻译器了。但是他还是想考考精通编程的你。

    BD语言只能处理32位有符号整数一种类型的数据(对应C++中的int类型)。在BD语言中,常量表示为“数字串*”,对应C++中的一个由该数字串表示的非负整数,其中数字串长度至少为1且允许有前导0(但翻译后需去掉前导0),该非负整数不能超过int范围;变量表示为“#数字串*”,对应C++的一个名字为“r数字串”的变量,该数字串长度同样至少为1。

     

    BD程序中有如下四种类型的语句(说明:在对应的C++语句中,下划线为空格):

    语句类型

    BD语言语句

    对应C++代码

    赋值语句

    1rxoy

    r_=_x_o_y;

    条件语句

    2xcy

    if_(x_c_y)_{

    循环语句

    3x

    for_(int_it_=_0;_it_<_x;_it++)_{

    结束语句

    0

    }

             其中,

    < >

    r是一个变量,x和y既可以是变量,也可以是常量;

    在赋值语句中,o是一个二元四则运算符,1为“+”,2为“-”,3为“*”,4为“/”。注意,若为除法,除数不能为常数0,但可以是变量

    在条件语句中,c是一个比较操作符,1为“==”,2为“<”,3为“>”;

    在循环语句对应的C++语句中,循环变量it的t是一个数字(从0开始编号),即循环变量可以是i0,i1,……,i9,i10,i11,……,在翻译时,应选择最小的不与外层循环相同的t。

    假设给定的BD程序为:3100*1#1*#1*11*1#0*#0*1#1*0

    步骤1

    根据之前给出的对应关系,可直译为:

    C++语句

    原BD语句

    for_(int_i0_=_0;_i0_<_100;_i0++)_{

    3100*

    r1_=_r1_+_1;

    1#1*#1*11*

    r0_=_r0_+_r1;

    1#0*#0*1#1*

    }

    0

    步骤2

    按如下方式在头尾增加相应内容,将其补充为完整的程序:

    int_main()_{

    int_r0_=_0;

    int_r1_=_0;

    for_(int_i0_=_0;_i0_<_100;_i0++)_{

    r1_=_r1_+_1;

    r0_=_r0_+_r1;

    }

    printf(“%d_%d”,_r0,_r1);

    return_0;

    }

    注意:

    < >

    在一开始按变量名字典序升序顺序定义程序中出现过的除循环变量以外的所有变量,并初始化为0;

    最后的输出语句按变量名字典序升序顺序输出除循环变量以外的所有变量;

    若没有需要输出的变量,则省略输出语句。

    步骤3

    度度熊不希望翻译出来的程序如此难看,于是你需要增加缩进:简单说来,就是如果上一行是以“{”结尾的,则从此所有行都增加4个空格的缩进;若当前行是一个“}”,则从此所有行都减少4个空格的缩进。

    经过此处理,程序变为:

    int_main()_{

    ____int_r0_=_0;

    ____int_r1_=_0;

    ____for_(int_i0_=_0;_i0_<_100;_i0++)_{

    ________r1_=_r1_+_1;

    ________r0_=_r0_+_r1;

    ____}

    ____printf(“%d_%d”,_r0,_r1);

    ____return_0;

    }

    步骤4

    如果代码块“{……}”中有且只有一条语句,将大括号省略。注意,

    < >

    空行及行末空格也应省略;

    main函数的大括号不能被省略;

    提示:if (...) {...}和for (...) {...}对于外层嵌套来说,应视为一条语句。

    输入
  • 仅有一行,为仅包含数字、“*”和“#”的字符串,即度度熊编写的BD程序,长度不超过500。
  • 输出
  • 第一行是一个整数,表示最长合法前缀的长度;
    从第二行起是经过翻译后的C++程序,要求完全按照上述格式,且不允许行末空格及多余空行。
  • 样例输入
  • 3100*1#1*#1*11*1#0*#0*1#1*0
  • 样例输出
  • 27
    int main() {
        int r0 = 0;
        int r1 = 0;
        for (int i0 = 0; i0 < 100; i0++) {
            r1 = r1 + 1;
            r0 = r0 + r1;
        }
        printf("%d %d", r0, r1);
        return 0;
    }
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2024年2月25日 22:38

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2024年2月25日 23:32

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2024年2月25日 23:33

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2024年2月25日 23:34

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2024年2月25日 23:35

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2024年2月26日 00:12

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2024年2月26日 00:25

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2024年2月26日 00:26

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2024年2月26日 00:30

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2024年2月26日 00:33

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2024年2月26日 00:33

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2024年2月26日 00:34

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