1 /* 把使用机会当成“花费” */
 2 # include <iostream>
 3 # include <cstdio>
 4 # include <cstring>
 5 # include <string>
 6 # include <cstdlib>
 7 # include <cmath>
 8 # include <ctime>
 9 # include <climits>
10 # include <memory>
11 # include <functional>
12 # include <algorithm>
13 # include <bitset>
14 # include <set>
15 # include <map>
16 # include <stack>
17 # include <vector>
18 # include <deque>
19 # include <queue>
20 # include <iomanip>
21 # include <utility>
22 using namespace std;
23 
24 # define lson l,m,rt<<1
25 # define rson r,m+1,rt<<1|1
26 # define lowbit(x) (x&(-x))
27 # define lcm(a,b) (a*b/__gcd(a,b))
28 typedef long long ll;
29 const ll mod=1e9+7;
30 const int maxn=1e6+50;
31 const double pi=acos(-1.0);
32 const int eps=1e-8;
33 
34 char s[205], t[1000005];
35 int dp[maxn][105];
36 //dp[i][j]表示t[1-i]和t[1-j]能比配需要的最小的k值
37 int main()
38 {
39     int n, m, k;
40     memset(dp, 125, sizeof(dp));
41     cin>>m>>n>>k;
42     scanf("%s", s+1);
43     scanf("%s", t+1);
44 
45     //如果模式串s是空串,则k为0
46     for(int i=0; i<=n; i++ )
47         dp[i][0] = 0;
48 
49     for(int i=1; i<=n; i++ )//t
50     {
51         for(int j=1; j<=m; j++ )//s
52         {
53             //s中前(j-1)个和t中前(i-1)个匹配后,s[j]和t[i]直接匹配
54             if( s[j]==t[i] )
55                 dp[i][j] = min(dp[i-1][j-1], dp[i][j]);//不需要”花费“
56 
57                 //在前面已经匹配的基础上,删掉t[i],此时s[j]依旧没有匹配上,所以是dp[i][j-1],而不是dp[i][j]
58                 dp[i][j-1] = min(dp[i][j-1], dp[i-1][j-1]+1);
59                 //在前面已经匹配的基础上,增加一个t[i]与s[j]匹配,但由于是虚拟的增加,所以仍是dp[i-1][j],而不是dp[i][j];
60                 dp[i-1][j] = min(dp[i-1][j], dp[i-1][j-1]+1);
61                 //在前已经匹配的基础上,换t[i]
62                 dp[i][j] = min(dp[i][j], dp[i-1][j-1]+1);
63         }
64     }
65 
66     for(int i=1; i<=n; i++ )
67     {
68         if( dp[i][m]<=k )
69         {
70             cout<<‘S‘<<endl;
71             return 0;
72         }
73     }
74     cout<<‘N‘<<endl;
75     return 0;
76 }