hdoj Simply Syntax 1433 (模拟)

Simply Syntax Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 409    Accepted Submission(s): 202


Problem Description In the land of Hedonia the official language is Hedonian. A Hedonian professor had noticed that many of her students still did not master the syntax of Hedonian well. Tired of correcting the many syntactical mistakes, she decided to challenge the students and asked them to write a program that could check the syntactical correctness of any sentence they wrote. Similar to the nature of Hedonians, the syntax of Hedonian is also pleasantly simple. Here are the rules: 

0. The only characters in the language are the characters p through z and N, C, D, E, and I. 

1. Every character from p through z is a correct sentence. 

2. If s is a correct sentence, then so is Ns. 

3. If s and t are correct sentences, then so are Cst, Dst, Est and Ist. 

4. Rules 0. to 3. are the only rules to determine the syntactical correctness of a sentence. 

You are asked to write a program that checks if sentences satisfy the syntax rules given in Rule 0. - Rule 4. 
 
Input The input consists of a number of sentences consisting only of characters p through z and N, C, D, E, and I. Each sentence is ended by a new-line character. The collection of sentences is terminated by the end-of-file character. If necessary, you may assume that each sentence has at most 256 characters and at least 1 character.
 
Output The output consists of the answers YES for each well-formed sentence and NO for each not-well-formed sentence. The answers are given in the same order as the sentences. Each answer is followed by a new-line character, and the list of answers is followed by an end-of-file character. 

 
Sample Input Cp Isz NIsz Cqpq  
Sample Output NO YES YES NO //题意： 判断一个句子是否是合格的有4个条件： 0、句子中只包含字母p---z和C,D,E,I,N； 1、p---z每个字母是一个句子（关键性条件）； 2、如果句子s是正确的，那么Ns是正确的； 3、如果句子s、t是正确的，那么Cst，Dst，Est,Ist是正确的； 4、判断句子正确的条件只有0--3. //思路：大神的思路，太牛了。。。

用一个cou表示当前有几个合法的句子 最后判断是不是仅有一个合法的句子。

 注意当搜索到N的时候cou一定要为1.否则break

如果搜索到CDEI的时候cou < 2 那么也要break

还有可能出现不是上述的字母也要break

#include <cstdio>
#include <cstring>
const int M = 300;
using namespace std;
char s[M];
int main()
{
	while(scanf("%s", s) == 1)
	{
		int len = strlen(s);
		int cou = 0;
		for(int i = len-1; i >= 0; -- i)
		{
			if(s[i] >= 'p'&&s[i] <= 'z')
			{
				++cou;
			}
			else if(s[i] == 'N')
			{
				if(cou == 0)
				{
					break;
				}
			}
			else if(s[i] == 'C'||s[i] == 'D'||s[i] == 'E'||s[i] == 'I')
			{
				if(cou >= 2) 
					--cou;
				else
				{
					cou = 0; 
					break;
				}
			}
			else 
			{
				cou = 0; 
				break;
			}
		}
		if(cou == 1) 
			printf("YES\n");
		else
			puts("NO");
	}
	return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 710
#define M 100000000
using namespace std;
char s[300];
int main()
{
	int i;
	while(gets(s)!=NULL)
	{
		int l=strlen(s);
		if(l==1)
		{
			if(s[0]>='p'&&s[0]<='z')
				printf("YES\n");
			else
				printf("NO\n");
		}
		else if(s[l-1]>='C'&&s[l-1]<='N')
			printf("NO\n");
		else
		{
			int flag=0;int kk=0;int k=0;
			for(i=l-1;i>=0;i--)
			{
				if(!((s[i]>='p'&&s[i]<='z')||s[i]=='C'||s[i]=='D'||s[i]=='E'||s[i]=='I'||s[i]=='N'))
				{
					flag=1;
					break;
				}
				if(s[i]>='p'&&s[i]<='z')
				{
					k++;
					continue;
				}
				else if(s[i]=='C'||s[i]=='D'||s[i]=='E'||s[i]=='I')
				{
					if(k>2||k+kk<2)
					{
						flag=1;
						break;
					}
					else	
					{
						k=1;
						if(kk>1)
							kk--;
					}
				}
				else if(s[i]=='N')
				{
					if(k>1)
					{
						flag=1;
						break;
					}
					else
					{
						if(k)
						{
							kk++;
							k=0;
						}	
					}
				}
			}
			if(k>1)
				flag=1;
			if(flag)
				printf("NO\n");
			else
				printf("YES\n");
		}
	}
	return 0;
}

Ns
Cpp
CpCpp
NCpp
CpCpp
CCpNs
N
I
p
z
NCpCpCpp
NNNNs
NCpNsCpp
NNNCpCpNNsNsCpp
ppppp
pppNs
PCpp
CINs
NICps
INsNs
NssCpp
INsNsNs
INsCpp
CpCpCpCpCpCp
CpCpCpp
CpNsNs
CpCNsNs
CpCNsNsNs
CNNs
CNNNNs
CppCpp
ICCCppNsNsNs//这两组数据错了， 
ICppNs