127.0.0.1:9200
URL: http://127.0.0.1:9200/likecs_art_db/_search
REQUEST:
Array
(
    [query] => Array
        (
            [match] => Array
                (
                    [text] => Array
                        (
                            [query] => bzoj 3744【Gty的妹子序列】
                        )

                )

        )

    [highlight] => Array
        (
            [fields] => Array
                (
                    [text] => stdClass Object
                        (
                        )

                )

            [pre_tags] => #em#
            [post_tags] => #/em#
        )

    [size] => 8
    [from] => 0
)
RESPONSE:
string(7818) "{"took":19,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":66.82138,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"48328","_score":66.82138,"_source":{"id":"48328","text":"bzoj 3744\u3010Gty\u7684\u59b9\u5b50\u5e8f\u5217\u3011","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"kgxw0430","tagsname":"\u5206\u5757\u5728\u79bb\u7ebf\u4e0e\u5728\u7ebf\u95ee\u9898\u7684\u5e94\u7528","tagsid":"[\"18692\"]","catesname":"\u6570\u636e\u7ed3\u6784\n                    ","catesid":"[\"6385\"]","createtime":"1547598342"},"highlight":{"text":["#em#bzoj#/em# #em#3744#/em#【#em#Gty#/em##em#的#/em##em#妹#/em##em#子#/em##em#序#/em##em#列#/em#】"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"14082","_score":23.673746,"_source":{"id":"14082","text":"bzoj 2423 \u6700\u957f\u516c\u5171\u5b50\u5e8f\u5217","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"kickit","tagsname":"\u6c42lcs\u4e2a\u6570","tagsid":"[7204]","catesname":"ACM-\u52a8\u6001\u89c4\u5212-\u7ebf\u6027dp|ACM-\u601d\u7ef4","catesid":"[2554,\"2548\"]","createtime":"1523522644"},"highlight":{"text":["#em#bzoj#/em# 2423 最长公共#em#子#/em##em#序#/em##em#列#/em#"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"10106","_score":20.205917,"_source":{"id":"10106","text":"BZOJ 5039: [Jsoi2014]\u5e8f\u5217\u7ef4\u62a4","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"Hammer-cwz-77","tagsname":"","tagsid":"[]","catesname":"\u7ebf\u6bb5\u6811","catesid":"[\"1340\"]","createtime":"1520772084"},"highlight":{"text":["#em#BZOJ#/em# 5039: [Jsoi2014]#em#序#/em##em#列#/em#维护"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"969513","_score":19.429579,"_source":{"id":"969513","text":"bzoj 1005 1211 prufer\u5e8f\u5217\u603b\u7ed3","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"CSU3901130321","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1639832534"},"highlight":{"text":["#em#bzoj#/em# 1005 1211 prufer#em#序#/em##em#列#/em#总结"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"897142","_score":18.906435,"_source":{"id":"897142","text":"\u6caa\u6f02\u7a0b\u5e8f\u5a9b\u59b9\u5b50\u7684\u4e00\u5929...","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"cxuanBlog","tagsname":null,"tagsid":"","catesname":"","catesid":"","createtime":"1638386847"},"highlight":{"text":["沪漂程#em#序#/em#媛#em#妹#/em##em#子#/em##em#的#/em#一天..."]}},{"_index":"likecs_art_db","_type":"_doc","_id":"969515","_score":18.757704,"_source":{"id":"969515","text":"[HNOI2008][bzoj 1005]\u660e\u660e\u7684\u70e6\u607c\uff08prufer\u5e8f\u5217\uff09","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"leom10","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1639832537"},"highlight":{"text":["[HNOI2008][#em#bzoj#/em# 1005]明明#em#的#/em#烦恼（prufer#em#序#/em##em#列#/em#）"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"48376","_score":18.33127,"_source":{"id":"48376","text":"bzoj 2038\u3010\u5c0fZ\u7684\u889c\u5b50\u3011","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"kgxw0430","tagsname":"\u5206\u5757\u4e0e\u83ab\u961f\u7b97\u6cd5","tagsid":"[18745]","catesname":"\u6570\u636e\u7ed3\u6784\n                    ","catesid":"[\"6385\"]","createtime":"1547610055"},"highlight":{"text":["#em#bzoj#/em# 2038【小Z#em#的#/em#袜#em#子#/em#】"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"969510","_score":17.548588,"_source":{"id":"969510","text":"BZOJ 1005 \u660e\u660e\u7684\u70e6\u607c(prufer\u5e8f\u5217+\u9ad8\u7cbe\u5ea6)","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"lishiyao","tagsname":"","tagsid":"","catesname":"","catesid":"","createtime":"1639832530"},"highlight":{"text":["#em#BZOJ#/em# 1005 明明#em#的#/em#烦恼(prufer#em#序#/em##em#列#/em#+高精度)"]}}]}}"
127.0.0.1:9200
URL: http://127.0.0.1:9200/likecs_art_db/_search
REQUEST:
Array
(
    [query] => Array
        (
            [match] => Array
                (
                    [text] => Array
                        (
                            [query] => bzoj 3744【Gty的妹子序列】
                        )

                )

        )

    [highlight] => Array
        (
            [fields] => Array
                (
                    [text] => stdClass Object
                        (
                        )

                )

            [pre_tags] => #em#
            [post_tags] => #/em#
        )

    [size] => 8
    [from] => 8
)
RESPONSE:
string(7601) "{"took":19,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":66.82138,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"137773","_score":17.546574,"_source":{"id":"137773","text":"7.29\u8003\u8bd5\u603b\u7ed3(NOIP\u6a21\u62df27)[\u725b\u534a\u4ed9\u7684\u59b9\u5b50\u56fe\u00b7Tree\u00b7\u5e8f\u5217]","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"Varuxn","tagsname":"","tagsid":"[]","catesname":"\u603b\u7ed3","catesid":"[\"736\"]","createtime":"1627562205"},"highlight":{"text":["7.29考试总结(NOIP模拟27)[牛半仙#em#的#/em##em#妹#/em##em#子#/em#图·Tree·#em#序#/em##em#列#/em#]"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"64027","_score":17.518856,"_source":{"id":"64027","text":"BZOJ 3990 \u6392\u5e8f","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"Juve","tagsname":"","tagsid":"[]","catesname":"\u8003\u8bd5","catesid":"[\"7919\"]","createtime":"1562660989"},"highlight":{"text":["#em#BZOJ#/em# 3990 排#em#序#/em#"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"337805","_score":17.518856,"_source":{"id":"337805","text":"bzoj AC\u5012\u5e8f","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        ","username":"fighting-to-the-end","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1636557769"},"highlight":{"text":["#em#bzoj#/em# 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bzoj 3744【Gty的妹子序列】 - 爱码网

bzoj 3744【Gty的妹子序列】

kgxw0430

Description

　　给定一个长度为 n 的序列，第 i 个位置上的数为 ai。现在有 m 组询问， 每组询问给定两个参数 L,R，要求输出从 i∈ [L,R] 中随机抽取两个数， 满足这两个数权值相同的概率。

　　n,m≤ 5×10^4,ai ≤n

分析

　　 这道题多组询问，求任意区间逆序对数，最暴力写法就是开n个树状数组，查询的时候利用前缀和的思路求一下逆序对数。但是这样好像还不如直接对于每次询问建立一个树状数组直接求来的好。（反正都是暴力，你怎么舒服怎么写呗）。这种题离线的话应该可以写，但它强制在线。考虑分块能否维护信息？

　　如果一道题可以写分块，套路是一样的，你考虑每一块要维护什么，边缘部分对答案的贡献怎么处理。对于这个题来说，显然你每一个块要维护的是逆序对数，边缘部分对答案的贡献显然是左边的与块内的数形成的逆序对数+左边的与右边的数形成的逆序对数+块内的数与右边的数形成的逆序对数。至于逆序对数怎么求，显然树状数组求，但对于任意边界划分的块都维护一个树状数组，显然不管是空间还是时间都说不过去。这时候就利用到了前缀和的思路，我们只开T个关于后缀的树状数组，求一个数在某个连续块内有多少个小于它的数的时候可以两个树状数组减一下。那么此时查询复杂度是m*s*log n，如果预处理每个块答案复杂度是暴力n^2/s^2*n log n的话，不管怎么调，都超出了1e8. 其实仔细想一下，我们处理每一个后缀的树状数组的时候可以求出T个块的答案。那么预处理复杂度就是n/s*n logn。调和一下，取s=sqrt（n），总复杂度是n sqrt（n）log n，足以通过此题。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int N=5e4+10;
int lan,t,n,m,p,q,s,a[N],b[300],c[300][N],sum[90000],d[N],c1[N];
void discrete(){
    sort(d+1,d+n+1);
    int sz=unique(d+1,d+n+1)-d-1;
    for(int i=1;i<=n;++i) a[i]=lower_bound(d+1,d+sz+1,a[i])-d;
}
void add(int id,int x){
    for(;x<=n;x+=x&(-x)) c[id][x]+=1;
}
int ask(int id,int x){
    int res=0;
    for(;x>0;x-=x&(-x)) res+=c[id][x];
    return res;
}
void add1(int x,int val){
    for(;x<=n;x+=x&(-x)) c1[x]+=val;
}
int ask1(int x){
    int res=0;
    for(;x>0;x-=x&(-x)) res+=c1[x];
    return res;
}
int work(int l,int r){
    l=l^lan,r=r^lan;
    if(l>r) swap(l,r);
    p=l/s,q=r/s;
    if(l%s==1) p++;
    else if(l%s) p+=2;
    else p++;
    if(q-p<0){
        int ans=0;
        for(int i=l;i<=r;++i){
            ans+=ask1(n)-ask1(a[i]);
            add1(a[i],1);
        }
        for(int i=l;i<=r;++i) add1(a[i],-1);
        return ans;
    }else{
        int pos=b[p-1]+q-p+1;
        int ans=sum[pos];
        for(int i=l;i<=s*(p-1);++i){
            ans+=ask1(n)-ask1(a[i]);
            add1(a[i],1);
        }
        for(int i=q*s+1;i<=r;++i){
            ans+=ask1(n)-ask1(a[i]);
            add1(a[i],1);
        }
        for(int i=l;i<=s*(p-1);++i){
            add1(a[i],-1);
        }
        for(int i=q*s+1;i<=r;++i){
            add1(a[i],-1);
        }
        for(int i=l;i<=s*(p-1);++i){
            ans+=ask(p,a[i]-1)-ask(q+1,a[i]-1);
        }
        for(int i=q*s+1;i<=r;++i){
            ans+=ask(p,n)-ask(p,a[i])-ask(q+1,n)+ask(q+1,a[i]);
        }
        return ans;
    }
}
int main(){
    //freopen("h.in","r",stdin);
    //freopen("h.out","w",stdout);
    scanf("%d",&n);
    t=sqrt(1.0*n);s=n/t;
    for(int i=1;i<=n;++i) scanf("%d",&a[i]),d[i]=a[i];
    for(int i=1;i<=t;++i) b[i]=b[i-1]+t-i+1;
    discrete();
    for(int i=1;i<=t;++i){
        int ans=0;
        for(int j=(i-1)*s+1;j<=n/s*s;++j){ //这个小错误搞了我一个晚上，这时候应该是分的块数乘上长度，如果直接写n，它多余的不足以构成一整块的还会影响答案         
　　　　　　　int pos=b[i-1]+j/s-i+1; //这个写的时候也得注意一下
            if(j%s) pos++;
            ans+=ask(i,n)-ask(i,a[j]);
            sum[pos]=ans;
            add(i,a[j]);
        }
    }
    scanf("%d",&m);
    for(int i=1;i<=m;++i){
        int l,r;scanf("%d%d",&l,&r);
        printf("%d\n",lan=work(l,r));
    }
    return 0;
}

 

 

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