HDU 4891 The Great Pan （模拟）

The Great Pan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 463    Accepted Submission(s): 179

Problem Description

As a programming contest addict, Waybl is always happy to take part in various competitive programming contests. One day, he was competing at a regional contest of Inventing Crappy Problems Contest(ICPC). He tried really hard to solve a "geometry" task without success.

After the contest, he found that the problem statement is ambiguous! He immediately complained to jury. But problem setter, the Great Pan, told him "There are only four possibilities, why don't you just try all of them and get Accepted?

".

Waybl was really shocked. It is the first time he learned that enumerating problem statement is as useful as trying to solve some ternary search problem by enumerating a subset of possible angle!

Three years later, while chatting with Ceybl, Waybl was told that some problem "setters" (yeah, other than the Great Pan) could even change the whole problem 30 minutes before the contest end! He was again shocked.

Now, for a given problem statement, Waybl wants to know how many ways there are to understand it.

A problem statement contains only newlines and printable ASCII characters (32 ≤ their ASCII code ≤ 127) except '{', '}', '|' and '$'.

Waybl has already marked all ambiguity in the following two formats:

1.{A|B|C|D|...} indicates this part could be understand as A or B or C or D or ....
2.$blah blah$ indicates this part is printed in proportional fonts, it is impossible to determine how many space characters there are.

Note that A, B, C, D won't be duplicate, but could be empty. (indicate evil problem setters addedclarified it later.)

Also note that N consecutive spaces lead to N+1 different ways of understanding, not 2^N ways.

It is impossible to escape from "$$" and "{}" markups even with newlines. There won't be nested markups, i.e. something like "${A|B}$" or "{$A$|B}" or "{{A|B}|C}" is prohibited. All markups will be properly matched.

 

Input

Input contains several test cases, please process till EOF.
For each test case, the first line contains an integer n, indicating the line count of this statement. Next n lines is the problem statement.
1 ≤ n ≤ 1000, size of the input file will not exceed 1024KB.

 

Output

For each test case print the number of ways to understand this statement, or "doge" if your answer is more than 10⁵.

 

Sample Input

9
I'll shoot the magic arrow several
 times on the ground, and of course
 the arrow will leave some holes
 on the ground. When you connect
 three holes with three line segments,
 you may get a triangle.
{|It is hole! Common sense!|
No Response, Read Problem
 Statement|don't you know what a triangle is?}
1
Case $1: = >$
5
$/*This is my code printed in
 proportional font, isn't it cool?
*/
printf("Definitely it is cooooooool 
%d
",4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 
* 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4);$
2
$Two  space$ and {blue|
red} color!

 

Sample Output

4
4
doge
6

 

题意：

题目给你n行文章，有两种情况，例如以下：

1.对于{A|B|C|D|...}，读出竖线的数量 x ，值为 x + 1

2.对于$blah blah$。读出每小段连续的空格数y1、y2...，值为(y1 + 1) * (y2 + 1) * ....

将以上求得的全部值相乘就可以。

思路：

模拟一遍就可以。

注意点：

1.注意数据相乘的积可能会超出int32位。甚至是long long的64位。

例如以下例子：

1
$ a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a$

一行共同拥有63个空。就会爆 long long。

解决方式是当超出 10^5 时，将值改为 10^5 + 1 ，而且不再处理字符串，详见代码。

/*************************************************************************
	> File Name: 1005.cpp
	> Author: Bslin
	> Mail: Baoshenglin1994@gmail.com
	> Created Time: 2014年07月29日 星期二 16时42分00秒
 ************************************************************************/

#include <cstdio>
#include <string.h>
using namespace std;
#define N 3050000

char str[N];

int main(int argc, char *argv[]) {
	freopen("in.txt", "r", stdin);
	int n, len, i, j, flagk, flag$, doge;
	long long space, ans, tmp1, tmp2;
	char nowk;
	while(scanf("%d", &n) != EOF) {
		ans = 1;
		tmp1 = 1;  // 竖杠数
		tmp2 = 1;  // 每段$$中的空格总乘积
		space = 1; // 每小段空格数
		flagk = flag$ = 0;  // 是否有"{"或"$"
		doge = 0;
		getchar();
		for (i = 0; i < n; ++i) {
			gets(str);
			len = strlen(str);
			if(doge) continue;
			for (j = 0; j < len; ++j) {
				if(str[j] == '{') {
					flagk = 1;
				} else if(str[j] == '}') {
					ans = ans * tmp1;
					tmp1 = 1;
					flagk = 0;
				} else if(str[j] == '$') {
					if(flag$ == 0) {
						flag$ = 1;
					} else {
						if(nowk == ' ') {
							tmp2 = tmp2 * space;
							if(tmp2 > 100000) tmp2 = 100001;   /////
						}
						ans = ans * tmp2;
						tmp2 = 1;
						space = 1;
						flag$ = 0;
					}
				} else {
					if(flagk) {
						if(str[j] == '|') {
							tmp1 ++;
						}
					}
					if(flag$) {
						if(str[j] == ' ') {
							space ++;
						} else {
							tmp2 = tmp2 * space;
							if(tmp2 > 100000) tmp2 = 100001;  /////
							space = 1;
						}
					}
				}
				if(ans > 100000) doge = 1;
				nowk = str[j];
			}
		}
		if(doge) {
			printf("doge
");
		} else {
			printf("%I64d
", ans);
		}
	}
	return 0;
}

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