uva 10340 All in All

解题思路：贪心

 1 /**************************************************************************
 2 user_id: SCNU20102200088
 3 problem_id: uva 10340
 4 problem_name: All in All
 5 **************************************************************************/
 6 
 7 #include <algorithm>
 8 #include <iostream>
 9 #include <iterator>
10 #include <iomanip>
11 #include <sstream>
12 #include <fstream>
13 #include <cstring>
14 #include <cstdlib>
15 #include <climits>
16 #include <bitset>
17 #include <string>
18 #include <vector>
19 #include <cstdio>
20 #include <cctype>
21 #include <ctime>
22 #include <cmath>
23 #include <queue>
24 #include <stack>
25 #include <list>
26 #include <set>
27 #include <map>
28 using namespace std;
29 
30 //线段树
31 #define lson l,m,rt<<1
32 #define rson m+1,r,rt<<1|1
33 
34 //手工扩展栈
35 #pragma comment(linker,"/STACK:102400000,102400000")
36 
37 const double EPS=1e-9;
38 const double PI=acos(-1.0);
39 const double E=2.7182818284590452353602874713526;  //自然对数底数
40 const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)
41 
42 const int x4[]={-1,0,1,0};
43 const int y4[]={0,1,0,-1};
44 const int x8[]={-1,-1,0,1,1,1,0,-1};
45 const int y8[]={0,1,1,1,0,-1,-1,-1};
46 
47 typedef long long LL;
48 
49 typedef int T;
50 T max(T a,T b){ return a>b? a:b; }
51 T min(T a,T b){ return a<b? a:b; }
52 T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
53 T lcm(T a,T b){ return a/gcd(a,b)*b; }
54 
55 ///////////////////////////////////////////////////////////////////////////
56 //Add Code:
57 ///////////////////////////////////////////////////////////////////////////
58 
59 int main(){
60     std::ios::sync_with_stdio(false);
61     //freopen("in.txt","r",stdin);
62     //freopen("out.txt","w",stdout);
63     ///////////////////////////////////////////////////////////////////////
64     //Add Code:
65     char a[100005],b[100005];
66     while(scanf("%s%s",a,b)!=EOF){
67         int alen=strlen(a),blen=strlen(b);
68         if(blen<alen){
69             printf("No\n");
70             continue;
71         }
72         int cnt=0,num=0;
73         while(num<blen){
74             if(a[cnt]==b[num]) cnt++;
75             num++;
76         }
77         if(cnt==alen) printf("Yes\n");
78         else printf("No\n");
79     }
80     ///////////////////////////////////////////////////////////////////////
81     return 0;
82 }
83 
84 /**************************************************************************
85 Testcase:
86 Input:
87 sequence subsequence
88 person compression
89 VERDI vivaVittorioEmanueleReDiItalia
90 caseDoesMatter CaseDoesMatter
91 Output:
92 Yes
93 No
94 Yes
95 No
96 **************************************************************************/

posted on 2013-09-27 17:40  SCNU20102200088  阅读(231)  评论(0)  编辑  收藏  举报

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