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：

/*
ID: asif
LANG: C++
TASK: test
*/
//# pragma comment(linker, "/STACK:102400000,102400000")
# include<iostream>
# include<cstdio>
# include<cstdlib>
# include<cstring>
# include<algorithm>
# include<cctype>
# include<cmath>
# include<string>
# include<set>
# include<map>
# include<stack>
# include<queue>
# include<vector>
# include<numeric>
using namespace std;
const int maxn=1111;
const double inf=0.000001;
const int INF=~0U>>1;
const int mod=1000000007;
# define lson l,m,rt<<1
# define rson m+1,r,rt<<1 | 1
# define PS printf("
")
# define S(n) scanf("%d",&n)
# define P(n) printf("%d
",n)
# define Ps(n) printf(" %d",(n))
# define SB(n) scanf("%lld",&n)
# define PB(n) printf("%lld
",n)
# define PBs(n) printf(" %lld",n)
# define SD(n) scanf("%lf",&n)
# define PD(n) printf("%.3lf
",n)
# define Sstr(s) scanf("%s",s)
# define Pstr(s) printf("%s
",s)
# define S0(a) memset(a,0,sizeof(a))
# define S1(a) memset(a,-1,sizeof(a))
typedef long long ll;
char str[maxn];
int a[6];
int main()
{
    //freopen("input.in", "r", stdin);
    //freopen("output.out", "w", stdout);
    while(~Sstr(str))
    {
        int l=strlen(str);
        S0(a);
        for(int i=0;i<l;i++)
        {
            if(str[i]=='n')
                a[0]++;
            else if(str[i]=='e')
                a[1]++;
            else if(str[i]=='w')
                a[2]++;
            else
                a[3]++;
        }
        if((a[0]&&a[1]&&a[2]&&a[3])||(a[0]&&a[2]&&a[1]&&(a[0]+a[1]+a[2]>3)))
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}