nbuoj 1532 New String

[1532] New String
时间限制: 2000 ms　内存限制: 65535 K
问题描述
Ceil so love the new string.
If you can erase any character from the string, and use the remain character by any order to comprise the "new" , the string is the new string, otherwise not.
Give you some string, you should help Ceil determine whether the string is the new string.
输入
There are multiple test cases. For each test case:
There is a string only contain the lowercase (1 <= string's length <= 1000).
输出
If the string is the new string ,output "YES", otherwise output "NO".
样例输入
```
aaneaw
nneeaa
wanaaea
```
样例输出
```
YES
NO
YES
```

思路：这是这场比赛里面最简单的一道题了，题意就是给你一个字符串，去掉一个字符，看看里面是否还同时存在'n','e','w'三个字符。PS：比　赛的时候第一次提交竟然没有加文件结束符号，返回一次TLE···nc啊

代码如下：

 1 /*
 2 ID: asif
 3 LANG: C++
 4 TASK: test
 5 */
 6 //# pragma comment(linker, "/STACK:102400000,102400000")
 7 # include<iostream>
 8 # include<cstdio>
 9 # include<cstdlib>
10 # include<cstring>
11 # include<algorithm>
12 # include<cctype>
13 # include<cmath>
14 # include<string>
15 # include<set>
16 # include<map>
17 # include<stack>
18 # include<queue>
19 # include<vector>
20 # include<numeric>
21 using namespace std;
22 const int maxn=1111;
23 const double inf=0.000001;
24 const int INF=~0U>>1;
25 const int mod=1000000007;
26 # define lson l,m,rt<<1
27 # define rson m+1,r,rt<<1 | 1
28 # define PS printf("
")
29 # define S(n) scanf("%d",&n)
30 # define P(n) printf("%d
",n)
31 # define Ps(n) printf(" %d",(n))
32 # define SB(n) scanf("%lld",&n)
33 # define PB(n) printf("%lld
",n)
34 # define PBs(n) printf(" %lld",n)
35 # define SD(n) scanf("%lf",&n)
36 # define PD(n) printf("%.3lf
",n)
37 # define Sstr(s) scanf("%s",s)
38 # define Pstr(s) printf("%s
",s)
39 # define S0(a) memset(a,0,sizeof(a))
40 # define S1(a) memset(a,-1,sizeof(a))
41 typedef long long ll;
42 char str[maxn];
43 int a[6];
44 int main()
45 {
46     //freopen("input.in", "r", stdin);
47     //freopen("output.out", "w", stdout);
48     while(~Sstr(str))
49     {
50         int l=strlen(str);
51         S0(a);
52         for(int i=0;i<l;i++)
53         {
54             if(str[i]=='n')
55                 a[0]++;
56             else if(str[i]=='e')
57                 a[1]++;
58             else if(str[i]=='w')
59                 a[2]++;
60             else
61                 a[3]++;
62         }
63         if((a[0]&&a[1]&&a[2]&&a[3])||(a[0]&&a[2]&&a[1]&&(a[0]+a[1]+a[2]>3)))
64             puts("YES");
65         else
66             puts("NO");
67     }
68     return 0;
69 }

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nbuoj 1532 New String

[1532] New String