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] => 2019 CCPC-Wannafly Winter Camp Day7(Div2, onsite)
                        )

                )

        )

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

                )

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

    [size] => 8
    [from] => 0
)
RESPONSE:
string(7491) "{"took":11,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":3851,"relation":"eq"},"max_score":101.33348,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"727285","_score":101.33348,"_source":{"id":"727285","text":"2019 CCPC-Wannafly Winter Camp Day7(Div2, onsite)","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":"mountaink","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1636133251"},"highlight":{"text":["#em#2019#/em# #em#CCPC#/em#-#em#Wannafly#/em# #em#Winter#/em# #em#Camp#/em# #em#Day7#/em#(#em#Div2#/em#, #em#onsite#/em#)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"52026","_score":73.97583,"_source":{"id":"52026","text":"CCPC-Wannafly Winter Camp Day1 (Div2, onsite)\u8865\u9898\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":"zookkk","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1550675304"},"highlight":{"text":["#em#CCPC#/em#-#em#Wannafly#/em# #em#Winter#/em# #em#Camp#/em# Day1 (#em#Div2#/em#, #em#onsite#/em#)补题总结"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"999806","_score":24.49516,"_source":{"id":"999806","text":"2017-2018 Petrozavodsk Winter Training Camp, Saratov SU Contest - NineSwords","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":"NineSwords","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1640494641"},"highlight":{"text":["2017-2018 Petrozavodsk #em#Winter#/em# Training #em#Camp#/em#, Saratov SU Contest - NineSwords"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"483535","_score":23.85418,"_source":{"id":"483535","text":"Petrozavodsk Winter-2019. Yandex Cup 2019.","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":"FST-stay-night","tagsname":"","tagsid":"","catesname":"","catesid":"","createtime":"1636177912"},"highlight":{"text":["Petrozavodsk #em#Winter#/em#-#em#2019#/em#. Yandex Cup #em#2019#/em#."]}},{"_index":"likecs_art_db","_type":"_doc","_id":"516410","_score":22.56801,"_source":{"id":"516410","text":"2019 China Collegiate Programming Contest Qinhuangdao Onsite","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":"Dup4","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1634555601"},"highlight":{"text":["#em#2019#/em# China Collegiate Programming Contest Qinhuangdao #em#Onsite#/em#"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"203507338","_score":21.890984,"_source":{"id":"203507338","text":"CS224n(winter 2019)\u7b14\u8bb0\u2014\u2014Lecture2","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":"","tagsname":null,"tagsid":"","catesname":null,"catesid":"","createtime":"1620950064"},"highlight":{"text":["CS224n(#em#winter#/em# #em#2019#/em#)笔记——Lecture2"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"672155","_score":20.20882,"_source":{"id":"672155","text":"2019 HZNU Winter Training Day 15 Comprehensive Training","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":"Tangent-1231","tagsname":"","tagsid":"","catesname":null,"catesid":"","createtime":"1635802673"},"highlight":{"text":["#em#2019#/em# HZNU #em#Winter#/em# Training Day 15 Comprehensive Training"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"76419","_score":19.749292,"_source":{"id":"76419","text":"2019-ccpc\u79e6\u7687\u5c9b\u73b0\u573a\u8d5b","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":"31415926535x","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1570274180"},"highlight":{"text":["#em#2019#/em#-#em#ccpc#/em#秦皇岛现场赛"]}}]}}"
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] => 2019 CCPC-Wannafly Winter Camp Day7(Div2, onsite)
                        )

                )

        )

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

                )

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

    [size] => 8
    [from] => 8
)
RESPONSE:
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#em#div2#/em#"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"356069","_score":16.078623,"_source":{"id":"356069","text":"2020 CCPC Weihai","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":"Lanly","tagsname":null,"tagsid":"","catesname":null,"catesid":"","createtime":"1632054052"},"highlight":{"text":["2020 #em#CCPC#/em# Weihai"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"18233","_score":16.007133,"_source":{"id":"18233","text":"CF#483(div2 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div2","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":"ckr1225","tagsname":"","tagsid":"[]","catesname":"\u7f16\u7a0b","catesid":"[\"118\"]","createtime":"1526214747"},"highlight":{"text":["bestcoder round 74 #em#div2#/em#"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"514664","_score":16.007133,"_source":{"id":"514664","text":"CodeForces Round 200 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127.0.0.1:9200
URL: http://192.168.101.128/searchcore/index.php/cihere_cn_db/_search
REQUEST:
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    [highlight] => Array
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                        (
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            [pre_tags] => #em#
            [post_tags] => #/em#
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    [from] => 0
)
RESPONSE:
bool(false)
127.0.0.1:9200
URL: http://127.0.0.1:9200/likecs_down_db/_search
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2019 CCPC-Wannafly Winter Camp Day7(Div2, onsite) - 爱码网

2019 CCPC-Wannafly Winter Camp Day7(Div2, onsite)

mountaink

solve 6/11

补题：

A.迷宫

Code:zz

Thinking:zz kk

把每个节点的深度都处理出来，同一深度的点的冲突度为 （x-1）,x为同层次点数减一。

然后冲突度不断下传（冲突度为3，则最多下传3层）

最后答案就是最后一层的深度加上冲突度。

#include<bits/stdc++.h> 
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=100010;
int n,m;
int a[maxn];
bool vis[maxn];
ll ans[maxn],No[maxn];
struct egde{
    int to,Next;
}e[maxn<<1];
struct node{
    int u;
    ll time;
};
int head[maxn],tot;
void init(){
    CLR(head,-1),tot=0;
    CLR(vis,0);
    CLR(ans,0);
    CLR(No,0);
}
void addv(int u,int v){
    e[++tot].to=v;
    e[tot].Next=head[u];
    head[u]=tot;
}
void bfs(){
    queue<node >q;
    q.push({1,0});
    vis[1] = true;
    ans[1] = 1;
    if(a[1])
    {
        No[1]++;
    }
    while(!q.empty())
    {
        node s=q.front();
        q.pop();
        int u=s.u;
        ll time=s.time;
        for(int i=head[u];i!=-1;i=e[i].Next)
        {
            int v=e[i].to;
            if(!vis[v])
            {
                vis[v] = true;
                q.push({v,time + 1});
                ans[v] = time + 1 + 1;
                if(a[v])
                {
                    No[time + 1 + 1]++;
                }
            }
        }
    }
}
int main(){
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        init();
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addv(u,v),addv(v,u);
        }
        bfs();
        ll aans = 0,mm = 0;
        for(int i = 1;i <= n;i++)
        {
            if(a[i])
            {
                mm = max(mm,ans[i]);
            }
        }
        for(int i = 1;i <= n;i++)
        {
            if(mm == i)
            {
                break;
            }
            No[i + 1] += max((ll)0,No[i] - 1);
        }
        printf("%lld\n",No[mm] - 1 + mm - 1);
    }
}
View Code

 

 

C.斐波那契数列

Code:pai爷  zz  kk

Thinking：pai爷  zz  kk

根据式子观察得，这个式子就是斐波那契数列丢掉自己二进制的低位1的前缀和，也就是 fib n - lowbit（fib n），所以暴力打表把减掉的这部分打出来，找规律，然后矩阵快速幂求斐波那契数列前n项和，减一减。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;
typedef long long ll;
const int p=998244353;
const int mod = 998244353;
ll f[1010],sum[1010],r;
int t;
void mul(ll f[3], ll a[3][3])
{
    ll c[3];
    memset(c, 0, sizeof(c));
    for (int j = 0; j < 3; j++)
        for (int k = 0; k < 3; k++)
            c[j] = (c[j] + (long long)f[k] * a[k][j]%mod) % mod;
    memcpy(f, c, sizeof(c));
}
void mulself(ll a[3][3])
{
    ll c[3][3];
    memset(c, 0, sizeof(c));
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++)
            for (int k = 0; k < 3; k++)
                c[i][j] = (c[i][j] + (long long)a[i][k] * a[k][j]%mod) % mod;
    memcpy(a, c, sizeof(c));
}
ll quickm(ll n){//求第n项和 
    ll f[3] = { 1,1,1 };
        ll tep=n-1;
        ll a[3][3] = { {1,1,1},{1,0,0},{0,0,1} };
        for (; tep; tep >>= 1)
        {
            if (tep & 1) mul(f, a);
            mulself(a);
        }
        return f[2];
}

ll solve(ll x,int w)
{
    if(x==0) return 0;
    if(x==1) return 1;
    if(x==2) return 2;
    if(x==3) return 4;
    if(x==4) return 5;
    if(x==5) return 6;
    if(x==6) return 14;
    for(int i=w;i>=1;i--)
    if(x>=1ll*6*f[i])
    {
        return (sum[i]+solve(x-6*f[i],i-1))%p;
    }
}
int main()
{
    f[1]=1;sum[1]=14;
    for(int i=2;i<=61;i++) 
    {
        f[i]=f[i-1]*2;
        sum[i]=(sum[i-1]*2%p+f[i]*4%p)%p;
    }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&r);
        ll ans=solve(r,61);
        //printf("%lld\n",ans);
        printf("%lld\n",(quickm(r)-ans+p)%p);
    }
}
View Code

 

D 二次函数

待补题 初中数学  

 

E 线性探查法

Code:kk

Thinking:kk

先把每个b[ i ]的余数求出来，余数等于i的说明没有冲突，不等于i的，假设偏移量为px，说明他的前px项都在他前面，所以就得到拓扑序，由于要字典序最小，所以用优先队列求拓扑序列。

div2暴力建边即可，div1要线段树优化建边？待学习。

#include<bits/stdc++.h> 
#include<unordered_map>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=1010;
int n;
ll a[maxn],b[maxn],m[maxn];
int dg[maxn];

struct egde{
    int to,Next;
}e[maxn*maxn];
int head[maxn],tot;
void init(){
    CLR(head,-1),tot=0;
}
void addv(int u,int v){
    e[++tot].to=v;
    e[tot].Next=head[u];
    head[u]=tot;
}
struct node{
    int pos;
    ll val;
    friend bool operator <(const node &a,const node &b){
        return a.val>b.val;
    }
};
priority_queue<node >q;
void Top(){
    int top=0;
    while(!q.empty()){
        node s=q.top();
        q.pop();
        a[++top]=s.val;
        for(int i=head[s.pos];i!=-1;i=e[i].Next)
        {
            int pos=e[i].to;
            dg[pos]--;
            if(dg[pos]==0)
            {
                q.push({pos,b[pos]});
            }
        }
    }
}
int main(){
    while(cin>>n)
    {
        init();
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&b[i]);
            m[i]=b[i]%n;
        }
        for(int i=0;i<n;i++)
        {
            if(m[i]==i)
            {
                q.push({i,b[i]});
            }else{
                int px=(i-m[i]+n)%n;
                while(px>0)
                {
                    addv((i-px+n)%n,i);
                    dg[i]++;
                    px--;
                }
            }
        }
        Top();
        for(int i=1;i<=n;i++)
        {
            printf("%lld%c",a[i],(i<n)?\' \':\'\n\');
        }
    }
}
View Code

 

F 逆序对!

Code：pai爷

Thinking：pai爷

0(n*n)枚举每个数对有贡献的答案为不相同的最高位。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
using namespace std;
const int p=998244353;
int n,m,x,f[1010][50];
int sum,ans=0,chu,yu,a,b;
int main()
{
//    freopen("1.txt","r",stdin);
//    freopen("1.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        int l=0;
        while(x>0)
        {
            f[i][++l]=x%2;
            x/=2;
        }
        f[i][0]=l;
    }
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        {
            int k;
            for(k=max(f[j][0],f[i][0]);k>=1;k--)
                if(f[i][k]!=f[j][k]) break;
                
            chu=m/(1<<(k-1));yu=m%(1<<(k-1));
            a=(chu-chu/2)*(1<<(k-1))-1;b=chu/2*(1<<(k-1));
            
            sum=yu+1;
            //printf("chu=%d yu=%d a=%d b=%d sum=%d k=%d\n",chu,yu,a,b,sum,k); 
            if(chu%2==0) a+=sum,b=m-a;
            else b+=sum,a=m-b;
            if(f[i][k]==0) ans=(ans+b)%p;
            else ans=(ans+a)%p;
            ans=(ans+p)%p;
            //printf("%d\n",ans);
        }
    printf("%d\n",ans);
}
View Code

 

G 抢红包机器人

Code：zz

Thinking：zz

暴力枚举每个人是机器人的情况，再dfs查找出所有的机器人，选取机器人最少的情况。

 

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
#include<stack>
#include<bitset>
#include<unordered_map>
const int maxn = 0x3f3f3f3f;
const double EI = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354594571382178525166427;
const double PI = 3.141592653589793238462643383279;
using namespace std;
struct s
{
    int k;
    int c[110];
}z[110];
vector<int>ve[110];
int vis[110];
unordered_map<int,int>mp[110];
inline void dfs(int pos)
{
    int si,i;
    si = ve[pos].size();
    for(i = 0;i < si;i++)
    {
        if(!vis[ve[pos][i]])
        {
            vis[ve[pos][i]] = 1;
            dfs(ve[pos][i]);
        }
    }
}
int main(void)
{
    //ios::sync_with_stdio(false);
    int n,m,i,j,ans,mi,l;
    while(~scanf("%d %d",&n,&m))
    {
        for(i = 0;i <= n;i++)
        {
            ve[i].clear();
            mp[i].clear();
        }
        for(i = 0;i < m;i++)
        {
            scanf("%d",&z[i].k);
            for(j = 0;j < z[i].k;j++)
            {
                scanf("%d",&z[i].c[j]);
            }
        }
        for(i = 0;i < m;i++)
        {
            for(j = z[i].k - 1;j >= 0;j--)
            {
                for(l = j - 1;l >= 0;l--)
                {
                    if(!mp[z[i].c[j]][z[i].c[l]])
                    {
                        mp[z[i].c[j]][z[i].c[l]] = 1;
                        ve[z[i].c[j]].push_back(z[i].c[l]);
                    }
                }
            }
        }
        ans = 100000;
        for(i = 1;i <= n;i++)
        {
            memset(vis,0,sizeof(vis));
            vis[i] = 1;
            dfs(i);
            mi = 0;
            for(j = 1;j <= n;j++)
            {
                if(vis[j])
                {
                    mi++;
                }
            }
            ans = min(ans,mi);
        }
        printf("%d\n",ans);
    }
    return 0;
}
View Code

 

H 同构

求补图，然后dp？待补

 

J 强壮的排列

Code：pai爷

Thinking：pai爷

暴力打表？待补

 

赛后总结

kk：今天是演员的一天，演队友演自己，开局A题就想错了思路，然后一直想错，两小时才把A题这水题做了，然后E题基本读完题意，稍微写写就得到正解了，结果建边数组少开了，re一发，给队友写个矩阵快速幂的板子，没注意数据范围，long long写成了int，tle一发。影帝影帝，以后演技要好一些，要注意细节。

pai爷：模数复制粘贴错误找了一个小时的bug而且还wa了三发。找规律速度不够快。

zz：这场比赛演技越来越好了，水题想了好多假算法，而且wa以后才发现。找规律的题贡献了一个表。

 

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