[2009国家集训队]小Z的袜子(hose)（BZOJ2038+莫队入门题）

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=2038

题目：

题意：中文题意，大家都懂。

思路：莫队入门题。不过由于要去概率，所以我们假设询问区间内有k中物品，每种物品我们假设它的数量为pi。那么我们可以进行下面一系列的公式推导：

所以我们用莫队维护该区间内某颜色的平方和，对于最简分式我们用一个gcd即可求出。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define bug printf("*********\n");
 #define FIN freopen("in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = 1e9 + ;
 const int maxn = 5e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 int n, m, block;
 ll sum;
 int a[maxn], cnt[maxn];

 struct node {
     int l, r, id;
     ll ans1, ans2;
     bool operator < (const node& x) const {
         return (l - ) / block == (x.l - ) / block ? r < x.r : (l - ) / block < (x.l - ) / block;
     }
 }ask[maxn];

 void update (int p, int val) {
     sum -= cnt[p] * cnt[p];
     cnt[p] += val;
     sum += cnt[p] * cnt[p];
 }

 ll gcd(ll a, ll b) {
     return b ==  ? a : gcd(b, a % b);
 }

 int main() {
     //FIN;
     while(~scanf("%d%d", &n, &m)) {
         block = sqrt(n);
         for(int i = ; i <= n; i++) {
             scanf("%d", &a[i]);
         }
         for(int i = ; i <= m; i++) {
             scanf("%d%d", &ask[i].l, &ask[i].r);
             ask[i].id = i;
         }
         sort(ask + , ask + m + );
         for(int i = , l = , r = ; i <= m; i++) {
             for(; r < ask[i].r; r++) update(a[r + ], );
             for(; r > ask[i].r; r--) update(a[r], -);
             for(; l < ask[i].l; l++) update(a[l], -);
             for(; l > ask[i].l; l--) update(a[l - ], );
             if(ask[i].l == ask[i].r) {
                 ask[ask[i].id].ans1 = , ask[ask[i].id].ans2 = ;
                 continue;
             }
             ll k1 = sum - (ask[i].r - ask[i].l + );
             ll k2 = (long long) (ask[i].r - ask[i].l + ) * (ask[i].r - ask[i].l);
             ll gg = gcd(k1, k2);
             ask[ask[i].id].ans1 = k1 / gg;
             ask[ask[i].id].ans2 = k2 / gg;
         }
         for(int i = ; i <= m; i++) {
             if(ask[i].ans1 == ) {
                 printf("0/1\n");
             } else {
                 printf("%lld/%lld\n", ask[i].ans1, ask[i].ans2);
             }
         }
     }
     return ;
 }