題意：在[l, r]之中任選兩個數，求它們相同的概率。

解：

莫隊入門。

概率這個很好搞，就是cnt * (cnt - 1) / 2。

然後發現每次挪指標的時候，某一個cnt會+1或-1。這時候差值就是2 * cnt_small。

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cmath>
 4 
 5 typedef long long LL;
 6 const int N = 50010;
 7 
 8 int a[N], bin[N], fr[N];
 9 LL ans;
 
10 
11 struct ASK {
12     int id, l, r;
13     LL ans;
14     inline bool operator <(const ASK &w) const {
15         if(fr[l] != fr[w.l]) {
16             return fr[l] < fr[w.l];
17         }
18         return r < w.r;
19     }
20 }ask[N];
21 
22 inline bool cmp(const ASK &a, const ASK &b) {
 
23     return a.id < b.id;
24 }
25 
26 inline LL gcd(LL a, LL b) {
27     if(!b) {
28         return a;
29     }
30     return gcd(b, a % b);
31 }
32 
33 inline void add(int x) {
34     ans += bin[a[x]] << 1;
35     bin[a[x]]++;
36     return;
37 }
38 
39 inline void del(int x) {
40     bin[a[x]]--;
 
41     ans -= bin[a[x]] << 1;
42 }
43 
44 int main() {
45     int n, m;
46     scanf("%d%d", &n, &m);
47     int T = sqrt(n);
48     for(int i = 1; i <= n; i++) {
49         scanf("%d", &a[i]);
50         fr[i] = (i - 1) / T + 1;
51     }
52     for(int i = 1; i <= m; i++) {
53         ask[i].id = i;
54         scanf("%d%d", &ask[i].l, &ask[i].r);
55     }
56     std::sort(ask + 1, ask + m + 1);
57 
58     int L = 1, R = 1;
59     bin[a[1]]++;
60     for(int i = 1; i <= m; i++) {
61         if(ask[i].l == ask[i].r) {
62             continue;
63         }
64         while(ask[i].l < L) {
65             add(--L);
66         }
67         while(R < ask[i].r) {
68             add(++R);
69         }
70         while(L < ask[i].l) {
71             del(L++);
72         }
73         while(ask[i].r < R) {
74             del(R--);
75         }
76         ask[i].ans = ans;
77     }
78 
79     std::sort(ask + 1, ask + m + 1, cmp);
80     for(int i = 1; i <= m; i++) {
81         if(!ask[i].ans || ask[i].l == ask[i].r) {
82             puts("0/1");
83             continue;
84         }
85         int g = gcd(ask[i].ans, 1ll * (ask[i].r - ask[i].l + 1) * (ask[i].r - ask[i].l));
86         printf("%lld/%lld\n", ask[i].ans / g, 1ll * (ask[i].r - ask[i].l + 1) * (ask[i].r - ask[i].l) / g);
87     }
88     return 0;
89 }