1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 const int maxn = 25;
 6 char s[maxn];
 7
 8 int main()
 9 {
10     scanf("%s", s);
11     int l = strlen(s);
12     for(int i = 0; i < l; ++i)
13     {
14         if(i == 0 && s[i] == '9') continue;
15         if(s[i] > '4') s[i] = '0' + '9' - s[i];
16     }
17     printf("%s\n", s);
18
19     return 0;
20 }

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 const int maxn = 1000 + 10;
 6
 7 int gcd(int a, int b)
 8 { return b == 0 ? a : gcd(b, a % b); }
 9
10 int main()
11 {
12     //freopen("in.txt", "r", stdin);
13
14     int n, x0, y0, cnt = 0;
15     scanf("%d%d%d", &n, &x0, &y0);
16     set<pair<int, int>> Set;
17     for(int i = 0; i < n; ++i)
18     {
19         int x, y;
20         scanf("%d%d", &x, &y);
21         x -= x0; y -= y0;
22         pair<int, int> t;
23         if(x == 0) t = make_pair(0, 1);//处于同一竖直线
24         else if(y == 0) t = make_pair(1, 0);//处于同一水平线
25         else
26         {
27             int g = gcd(x, y);
28             x /= g; y /= g;
29             if(x < 0) { x = -x; y = -y; }
30             t = make_pair(x, y);
31         }
32         if(!Set.count(t)) { cnt++; Set.insert(t); }
33     }
34
35     printf("%d\n", cnt);
36
37     return 0;
38 }

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4
 5 const LL MOD = 1000000000007LL;
 6
 7 inline LL Hash(string& s)
 8 {
 9     LL ans = 0;
10     for(int i = 0; i < s.length(); ++i)    ans = (ans*3+(s[i]-'a')) % MOD;
11     return ans;
12 }
13
14 int main()
15 {
16     //freopen("in.txt", "r", stdin);
17
18     int n, m;
19     scanf("%d%d", &n, &m);
20     set<LL> Set;
21     string s;
22     for(int i = 0; i < n; ++i)
23     {
24         cin >> s;
25         Set.insert(Hash(s));
26     }
27
28     for(int i = 0; i < m; ++i)
29     {
30         cin >> s;
31         bool flag = false;
32         LL x = Hash(s);//询问串的hash值
33         LL b = 1;
34         for(int j = s.length()-1; j >= 0; j--)
35         {
36             LL t = (x-b*(s[j]-'a'+1) % MOD + MOD) % MOD;//先将枚举的位置变为0
37             for(char c = 'a'; c <= 'c'; c++) if(c != s[j])
38             {
39                 LL tt = (t + b*(c-'a'+1)) % MOD;//新串对应的Hash值
40                 if(Set.count(tt)) { flag = true; break; }
41             }
42             if(flag) break;
43             b = (b*3) % MOD;//b = (3^i)%MOD
44         }
45
46         printf("%s\n", flag ? "YES" : "NO");
47     }
48
49     return 0;
50 }