C.0689-The 2019 ICPC China Shaanxi Provincial Programming Contest

We call a string as a 0689-string if this string only consists of digits '0', '6', '8' and '9'. Given a 0689-string $s$ of length $n$, one must do the following operation exactly once: select a non-empty substring of $s$ and rotate it 180 degrees.

More formally, let $s_i$ be the $i$-th character in string $s$. After rotating the substring starting from $s_l$ and ending at $s_r$ 180 degrees ($1 \le l \le r \le n$), string $s$ will become string $t$ of length $n$ extracted from the following equation, where $t_i$ indicates the $i$-th character in string $t$: $$t_i = \begin{cases} s_i & \text{if } 1 \le i < l \text{ or } r < i \le n \\ \text{'0'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'0'} \\ \text{'6'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'9'} \\ \text{'8'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'8'} \\ \text{'9'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'6'} \\ \end{cases}$$

What's the number of different strings one can get after the operation?

We hereby explain the first sample test case.

Substring	Result		Substring	Result
0	0689		68	0899
6	0989		89	0668
8	0689		068	8909
9	0686		689	0689
06	9089		0689	6890

It's easy to discover that we can get 8 different strings after the operation.

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题意：给定一个含有0689的串，你可以中心旋转子串，询问旋转任意子串产生的不同的串一共最多哟多少种？

我们假设从前往后统计，能么0和后面非0位置的串都可以统计，如果两个0之间有非0，能么会在非0的时候进行统计，这样就不重不漏了，同理8也是，但是6和9不行，9和6也不行，但是6和6以及自身还有9和9以及自身都是可以的

 1 #include <cstdio>
 2 #include <cstring>
 3
 4 const int MAXN = (int)1e6 + 5;
 5 int t;
 6 char str[MAXN];
 7 long long num[MAXN][5];
 8
 9 int main() {
10     scanf("%d", &t);
11     while (t--) {
12         scanf("%s", str + 1);
13         int n = strlen(str + 1);
14         for (int i = 1; i <= n + 1; i++) num[i][1] = num[i][2] = num[i][3] = num[i][4] = 0;
15         for (int i = n; i >= 1; i--) {
16             num[i][1] = num[i + 1][1] + (str[i] == '0');
17             num[i][2] = num[i + 1][2] + (str[i] == '8');
18             num[i][3] = num[i + 1][3] + (str[i] == '6');
19             num[i][4] = num[i + 1][4] + (str[i] == '9');
20         }
21         long long ans = 1;
22         for (int i = 1; i <= n; i++) {
23             if (str[i] == '0') {
24                 ans += num[i + 1][2];
25                 ans += num[i + 1][3];
26                 ans += num[i + 1][4];
27             }
28             if (str[i] == '8') {
29                 ans += num[i + 1][1];
30                 ans += num[i + 1][3];
31                 ans += num[i + 1][4];
32             }
33             if (str[i] == '6') {
34                 ans += num[i + 1][1];
35                 ans += num[i + 1][2];
36                 ans += num[i][3];
37             }
38             if (str[i] == '9') {
39                 ans += num[i + 1][1];
40                 ans += num[i + 1][2];
41                 ans += num[i][4];
42             }
43         }
44         if (num[1][3] == n || num[1][4] == n) ans--;
45         printf("%lld\n", ans);
46     }
47     return 0;
48 }