[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：

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 typedef long long ll;
17 typedef unsigned long long ull;
18 
19 #define bug printf("*********\n");
20 #define FIN freopen("in.txt", "r", stdin);
21 #define debug(x) cout<<"["<<x<<"]" <<endl;
22 #define IO ios::sync_with_stdio(false),cin.tie(0);
23 
24 const double eps = 1e-8;
25 const int mod = 1e9 + 7;
26 const int maxn = 5e4 + 7;
27 const double pi = acos(-1);
28 const int inf = 0x3f3f3f3f;
29 const ll INF = 0x3f3f3f3f3f3f3f3f;
30 
31 int n, m, block;
32 ll sum;
33 int a[maxn], cnt[maxn];
34 
35 struct node {
36     int l, r, id;
37     ll ans1, ans2;
38     bool operator < (const node& x) const {
39         return (l - 1) / block == (x.l - 1) / block ? r < x.r : (l - 1) / block < (x.l - 1) / block;
40     }
41 }ask[maxn];
42 
43 void update (int p, int val) {
44     sum -= cnt[p] * cnt[p];
45     cnt[p] += val;
46     sum += cnt[p] * cnt[p];
47 }
48 
49 ll gcd(ll a, ll b) {
50     return b == 0 ? a : gcd(b, a % b);
51 }
52 
53 int main() {
54     //FIN;
55     while(~scanf("%d%d", &n, &m)) {
56         block = sqrt(n);
57         for(int i = 1; i <= n; i++) {
58             scanf("%d", &a[i]);
59         }
60         for(int i = 1; i <= m; i++) {
61             scanf("%d%d", &ask[i].l, &ask[i].r);
62             ask[i].id = i;
63         }
64         sort(ask + 1, ask + m + 1);
65         for(int i = 1, l = 1, r = 0; i <= m; i++) {
66             for(; r < ask[i].r; r++) update(a[r + 1], 1);
67             for(; r > ask[i].r; r--) update(a[r], -1);
68             for(; l < ask[i].l; l++) update(a[l], -1);
69             for(; l > ask[i].l; l--) update(a[l - 1], 1);
70             if(ask[i].l == ask[i].r) {
71                 ask[ask[i].id].ans1 = 0, ask[ask[i].id].ans2 = 1;
72                 continue;
73             }
74             ll k1 = sum - (ask[i].r - ask[i].l + 1);
75             ll k2 = (long long) (ask[i].r - ask[i].l + 1) * (ask[i].r - ask[i].l);
76             ll gg = gcd(k1, k2);
77             ask[ask[i].id].ans1 = k1 / gg;
78             ask[ask[i].id].ans2 = k2 / gg;
79         }
80         for(int i = 1; i <= m; i++) {
81             if(ask[i].ans1 == 0) {
82                 printf("0/1\n");
83             } else {
84                 printf("%lld/%lld\n", ask[i].ans1, ask[i].ans2);
85             }
86         }
87     }
88     return 0;
89 }

 

posted @ 2018-07-26 21:21  Dillonh  阅读(275)  评论(0)  编辑  收藏  举报