T1：

　　考場上以為是數論題，想推性質，然而並不會，於是考慮60分

考慮發現數據範圍中k <= 40，考慮模數很小，那麼考慮計數問題的經典

解決方法——對映，考慮在1~k範圍內求解，在反射回原值域，範圍可以通過

不等式限制

　　正解並不是數論，而是資料結構維護，考慮類似排列，我們通過列舉

一維來解決限制，便於維護，通常以列舉中間量，可以限制出左右兩側變數

於是列舉b，發現一個很優秀的性質為a為一次，於是a + b^2的範圍即可確定

，那麼考慮c的貢獻（即落到多少區間內），邊更新邊計算即可，樹狀陣列

區間修改單點查詢

程式碼如下：

 1 #include <bits/stdc++.h>
 2
 
 using namespace std;
 3 #define I long long
 4 #define C char
 5 #define B bool
 6 #define V void
 7 #define D double
 8 #define LL long long
 9 #define UI unsigned int
10 #define UL unsigned long long
11 #define P pair
12 #define MP make_pair
13 #define a first
14 #define b second
15 #define debug cout << "It's Ok Here !" << endl;
16
 
 #define lowbit(x) (x & -x)
17 #define FP(x) freopen (#x,"r",stdin)
18 #define FC(x) freopen (#x,"w",stdout)
19 const I N = 1e5 + 3;
20 I T,n,t;
21 inline I read () {
22     I x(0),y(1); C z(getchar());
23     while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); }
24     while ( isdigit(z))  x = x * 10
 
 + (z ^ 48), z = getchar();
25     return x * y;
26 }
27 inline V Max (I &a,I b) { a = a > b ? a : b; }
28 inline V Min (I &a,I b) { a = a < b ? a : b; }
29 inline I max (I a,I b) { return a > b ? a : b; }
30 inline I min (I a,I b) { return a < b ? a : b; }
31 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; }
32 inline I abs (I a) { return a >= 0 ? a : -a; }
33 inline P<I,I> operator + (const P<I,I> &a,const P<I,I> &b) {
34     return MP (a.a + b.a,a.b + b.b);
35 }
36 inline P<I,I> operator - (const P<I,I> &a,const P<I,I> &b) {
37     return MP (a.a - b.a,a.b - b.b);
38 }
39 struct BIT {
40     I c[N];
41     inline V secmod (I x,I y) { 
42         for (;x <= t;x += lowbit (x))
43             c[x] ++ ;
44         for (;y <= t;y += lowbit (y))
45             c[y] -- ;
46     }
47     inline I preque (I x) { I ans(0);
48         for (; x ;x -= lowbit (x))
49             ans += c[x];
50         return ans;
51     }
52 }B1;    
53 signed main () {
54     FP (exclaim.in), FC (exclaim.out);
55     T = read();
56     for (I i(1);i <= T; ++ i) {
57         printf ("Case %d: ",i);
58         n = read(), t = read(); I tmp(0); LL ans(0);
59         for (I j(1);j <= n; ++ j) {
60             I l((j * j + 1) % t + 1), 
61               r((j * j + j) % t + 1);
62             if (j >= t) tmp += j / t - (r + 1 == l) - (r - l + 1 == t);
63             if (l <= r) B1.secmod (l,r + 1);
64             if (l >  r) B1.secmod (1,r + 1), B1.secmod (l,t + 1);
65             I last (ans);
66             ans += tmp + B1.preque (j * j * j % t + 1);
67         }
68         printf ("%lld\n",ans);
69         memset (B1.c,0,sizeof B1.c);
70     }
71 }

T2：

　　首先看錯題目暴斃，要求迴圈同構，於是首先簡單的想法為n^2列舉區間

那麼考慮如何O(1)判斷迴圈同構，字串匹配一種常見的方法顯然為雜湊

然而並不能應對迴圈，考慮剪枝，這裡應用到字符集優化，即通過字符集雜湊

首先判斷兩端字元是否相等，在進行迴圈判斷，可以通過此題，再通過map記錄

可以應對Hack資料

程式碼如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define I int
 4 #define C char
 5 #define B bool
 6 #define V void
 7 #define D double
 8 #define LL long long
 9 #define UI unsigned int
10 #define UL unsigned long long
11 #define P pair
12 #define MP make_pair
13 #define a first
14 #define b second
15 #define debug cout << "It's Ok Here !" << endl;
16 #define lowbit(x) (x & -x)
17 #define FP(x) freopen (#x,"r",stdin)
18 #define FC(x) freopen (#x,"w",stdout)
19 const I N = 5e3 + 3;
20 I n,ans;
21 UL p[N],f[N],t[N];
22 inline I read () {
23     I x(0),y(1); C z(getchar());
24     while (!isdigit(z)) { if (z == '-') y = -1; z = getchar(); }
25     while ( isdigit(z))  x = x * 10 + (z ^ 48), z = getchar();
26     return x * y;
27 }
28 inline V Max (I &a,I b) { a = a > b ? a : b; }
29 inline V Min (I &a,I b) { a = a < b ? a : b; }
30 inline I max (I a,I b) { return a > b ? a : b; }
31 inline I min (I a,I b) { return a < b ? a : b; }
32 inline V swap (I &a,I &b) { a ^= b, b ^= a, a ^= b; }
33 inline I abs (I a) { return a >= 0 ? a : -a; }
34 inline P<I,I> operator + (const P<I,I> &a,const P<I,I> &b) {
35     return MP (a.a + b.a,a.b + b.b);
36 }
37 inline P<I,I> operator - (const P<I,I> &a,const P<I,I> &b) {
38     return MP (a.a - b.a,a.b - b.b);
39 }
40 inline B Check (I l,I r) {
41     I mid (l + r >> 1), len (r - l + 1 >> 1);
42     if (t[r] - t[mid] != t[mid] - t[l - 1]) return false;
43     UL com (f[r] - f[mid] * p[len]);
44     for (I i(l);i <= mid; ++ i) 
45         if ((f[mid] - f[i] * p[mid - i] - f[l - 1]) * p[i - l + 1] + f[i] == com)
46             return true;
47     return false;
48 }
49 signed main () {
50     FP (s.in), FC (s.out);
51     n = read(); p[0] = 1; 
52     for (I i(1);i <= 5000; ++ i)
53         p[i] = p[i - 1] * 13331;
54     for (I i(1),x,y;i <= n; ++ i) {
55         x = read();
56         t[i] = t[i - 1] + p[x];
57         f[i] = f[i - 1] * 13331 + x;
58         if (x == y) ans ++ ;  y = x;
59     }
60     for (I len(3);len <  n;len += 2) 
61       for (I i(1);i + len <= n; ++ i) {
62         if (Check (i,i + len)) ans ++ ;
63       }
64     printf ("%d\n",ans);
65 }