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[1532] New String
- 时间限制: 2000 ms 内存限制: 65535 K
- 问题描述
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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
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There are multiple test cases. For each test case:
There is a string only contain the lowercase (1 <= string‘s length <= 1000). - 输出
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If the string is the new string ,output "YES", otherwise output "NO".
- 样例输入
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aaneaw nneeaa wanaaea
- 样例输出
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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("\n") 29 # define S(n) scanf("%d",&n) 30 # define P(n) printf("%d\n",n) 31 # define Ps(n) printf(" %d",(n)) 32 # define SB(n) scanf("%lld",&n) 33 # define PB(n) printf("%lld\n",n) 34 # define PBs(n) printf(" %lld",n) 35 # define SD(n) scanf("%lf",&n) 36 # define PD(n) printf("%.3lf\n",n) 37 # define Sstr(s) scanf("%s",s) 38 # define Pstr(s) printf("%s\n",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 }