BZOJ 2038 小Z的袜子(hose)（莫队算法）

题意:

中文题面

分析:

对于[L,R]的询问,设其中颜色为x,y,z,...的袜子的个数为a,b,c,...
那么ans(L,R)=C2a+C2b+C2c+...C2R−L+1=(a∗(a−1)/2+b∗(b−1)/2+c∗(c−1)/2+...)(R−L+1)∗(R−L)/2)
=a2+b2+c2+...−(a+b+c+...)(R−L+1)∗(R−L)=a2+b2+c2+...−(R−L+1)(R−L+1)∗(R−L)
所以这道题目的关键是求一个区间内每种颜色数目的平方和

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关于莫队具体的可以看这里, 戳我查看
莫队均摊复杂度为O(n1.5),证明如下:
对于每一块内的所有询问,r是单调递增的,总变化范围为0∼n,这部分总复杂度为O(n);
l的变化范围为0∼n−−√,每块内平均有n−−√个询问,这部分复杂度为O(n−−√⋅n−−√)=O(n)
总共有n−−√块询问,总体均摊复杂度为O(n−−√⋅(n+n))=O(2nn−−√)=O(n1.5)　　　■

代码(区间表示为[l,r) ):

////  Created by TaoSama on 2016-01-25//  Copyright (c) 2015 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 5e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;const int B = 250;typedef long long LL;int n, q;int c[N], cnt[N];LL sum, up[N], dw[N];struct Query {    int l, r, id;    bool operator< (const Query& q) const {        return r < q.r;    }};void update(int i, int delta) {    sum -= cnt[c[i]] * cnt[c[i]];    cnt[c[i]] += delta;    sum += cnt[c[i]] * cnt[c[i]];}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &n, &q) == 2) {        for(int i = 1; i <= n; ++i) scanf("%d", c + i);        memset(cnt, 0, sizeof cnt);        vector<vector<Query> > Q(n / B + 2, vector<Query>());        for(int i = 1; i <= q; ++i) {            int l, r; scanf("%d%d", &l, &r);            Q[l / B].push_back(Query {l, ++r, i});        }        for(int i = 0; i <= n / B; ++i) sort(Q[i].begin(), Q[i].end());        sum = 0;        for(int i = 0; i < Q.size(); ++i) {            int l, r; l = r = i * B;            for(int j = 0; j < Q[i].size(); ++j) {                Query &q = Q[i][j];                while(r < q.r) update(r++, 1);                while(l < q.l) update(l++, -1);                while(l > q.l) update(--l, 1);                up[q.id] = sum - r + l;                dw[q.id] = 1LL * (r - l) * (r - l - 1);                LL g = __gcd(up[q.id], dw[q.id]);                up[q.id] /= g;                dw[q.id] /= g;            }            for(int j = l; j < r; ++j) update(j, -1);        }        for(int i = 1; i <= q; ++i) printf("%lld/%lld\n", up[i], dw[i]);    }    return 0;}